Numeric rows: definitions, properties, signs of convergence, examples, solutions. The convergence of a series of online why a number 1 n diverge
You can check the convergence of a number in several ways. First, you can simply find the sum of the row. If as a result we get a finite number, then such a number converge. For example, since
then a series converges. If we failed to find the amount of a number, then other methods should be used to verify the convergence of the series.
One of these methods is sign of Dalamber
here and, accordingly, nth and (n + 1) -th members of the series, and convergence is determined by the value D: if D< 1 - ряд сходится, если D >
As an example, we investigate the convergence of a number using a sign of the Dalamber. First, write expressions for and. Now we find the appropriate limit:
Since, according to the sign of the Dalamber, the series converges.
Another method that allows you to check the convergence of the series is cauchy radical signwhich is written as follows:
here is a nth member of a series, and convergence, as in the case of a sign of the Dalamber, is determined by the value D: if D< 1 - ряд сходится, если D > 1 - diverge. When d \u003d 1 - this feature does not give an answer and additional research should be carried out.
As an example, we investigate the convergence of a series using a radical sign of Cauchy. First write the expression for. Now we find the appropriate limit:
Since Title \u003d "15625/64\u003e 1"\u003e, in accordance with the radical sign of Cauchy, a row diverges.
It is worth noting that along with those listed, there are other signs of convergence of rows, such as the integral sign of Cauchy, a sign of Raabe, etc.
Our online calculator based on the Wolfram Alpha system allows you to test the convergence of the row. At the same time, if the calculator gives a specific number as the sum of the row, then the series converges. Otherwise, it is necessary to pay attention to the item "Test of the convergence of the series". If there is a phrase "Series Converges", the series converges. If the phrase is present "Series Diverges", then a row diverges.
Below is the transfer of all possible values \u200b\u200bof the Point Test of the Row Test:
| English text | Text in Russian |
|---|---|
| By The Harmonic Series Test, The Series Diverges. | When comparing the series under study with a harmonic side, the initial range is diverged. |
| The Ratio Test Is inconclusive. | The sign of Dalamber cannot give a response about the convergence of the series. |
| THE ROOT TEST IS INCONCLUSIVE. | The radical sign of Cauchi cannot give a response about the convergence of the series. |
| By The Comparison Test, The Series Converges. | On the sign of comparison, the series converges |
| By The Ratio Test, The Series Converges. | On the sign of the Dalamber, the series converges |
| By The Limit Test, The Series Diverges. | On the main thing that title \u003d "(! Lang: the limit of the n-bous member of the series at n-\u003e oo is not equal to zero or does not exist"> , или указанный предел не существует, сделан вывод о том, что ряд расходится. !} |
Find the sum of the number of numbers. If it fails to find it, the system calculates the sum of a number with a certain accuracy.
The convergence of the row
This calculator can determine whether a series converges, also shows - what signs of convergence are triggered, and which is not.
Also knows how to determine the convergence of power rows.
A graph of a number is also built, where you can see the speed of the convergence of a series (or divergence).
Rules for entering expressions and functions
Expressions may consist of functions (designations are given in alphabetical order): absolute (X) Absolute value x.
(module x. or | X |)
arcCOS (X) Function - Arkkosinus from x. arccosh (x) Arkkosinus hyperbolic OT x. arcsin (X) Arksinus Ot x. arcsinh (x) Arksinus hyperbolic OT x. arctg (x) Function - Arctangent from x. arctgh (x) Arctangent hyperbolic OT x. e. e. The number that is approximately 2.7 exp (x) Function - Exhibitor from x. (that I. e.^x.)
log (x) Or. lN (X) Natural logarithm OT. x.
(To obtain log7 (x), you need to enter log (x) / log (7) (or, for example for log10 (x)\u003d log (x) / log (10)) pI The number is "Pi", which is approximately equal to 3.14 sIN (X) Function - sinus from x. cOS (X) Function - cosine from x. sINH (X) Function - sine hyperbolic from x. cOSH (X) Function - cosine hyperbolic from x. sQRT (X) Function - Square root from x. sQR (X) or x ^ 2. Function - Square x. tG (X) Function - Tangent from x. tGH (X) Function - Tangent hyperbolic from x. cBRT (X) Function - Cubic root from x.
The following operations can be applied in expressions: Actual numbers Enter in the form 7.5
, not 7,5
2 * X. - Multiplication 3 / X. - division x ^ 3. - Establishment x + 7. - Addition x - 6. - subtraction
Other features: floor (x) Function - rounding x. In a smaller side (Example Floor (4.5) \u003d\u003d 4.0) ceiling (X) Function - rounding x. In the most side (Example Ceiling (4.5) \u003d\u003d 5.0) sIGN (X) Function - sign x. eRF (X) Error function (or probability integral) laplace (X) Laplas function
Harmonic row - The amount composed of an infinite number of members inverse the sequential numbers of the natural range:
Σ k \u003d 1 ∞ 1 k \u003d 1 + 1 2 + 1 3 + 1 4 + ⋯ + 1 k + ⋯ (\\ displaystyle \\ sum _ (k \u003d 1) ^ (\\ mathcal (\\ infty)) (\\ FRAC (1 ) (k)) \u003d 1 + (\\ FRAC (1) (2)) + (\\ FRAC (1) (3)) + (\\ FRAC (1) (4)) + \\ CDOTS + (\\ FRAC (1) (k)) + \\ CDOTS).Encyclopedic YouTube.
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✪ Numeric rows. Basic Concepts - Bezbotvy
✪ Proof of the displacement of the harmonic series
✪ Numeric rows-9. Convergence and divergence of Dirichlet
✪ Consultation №1. Mat. analysis. Fourier row on trigonometric system. Simplest properties
✪ rows. Overview
Subtitles
The sum of the first n members of the series
Separate members of the row tend to zero, but its amount is diverged. The N-by the partial sum S n of the harmonic row is called a N-accomplished harmonic number:
Sn \u003d Σ k \u003d 1 n 1 k \u003d 1 + 1 2 + 1 3 + 1 4 + ⋯ + 1 n (\\ displaystyle s_ (n) \u003d \\ sum _ (k \u003d 1) ^ (n) (\\ FRAC (1 ) (k)) \u003d 1 + (\\ FRAC (1) (2)) + (\\ FRAC (1) (3)) + (\\ FRAC (1) (4)) + \\ CDOTS + (\\ FRAC (1) (n)))Some partial sums
| S 1 \u003d 1 S 2 \u003d 3 2 \u003d 1, 5 S 3 \u003d 11 6 ≈ 1,833 s 4 \u003d 25 12 ≈ 2,083 s 5 \u003d 137 60 ≈ 2,283 (\\ DisplayStyle (\\ Begin (Matrix) S_ (1) & \u003d & 1 \\\\\\\\ S_ (2) & \u003d & (\\ FRAC (3) (2)) & \u003d & 1 (,) 5 \\\\\\\\ S_ (3) & \u003d & (\\ FRAC (11) (6)) & \\ Approx & 1 (,) 833 \\\\\\\\ S_ (4) & \u003d & (\\ FRAC (25) (12)) \\ Approx & 2 (,) 083 \\\\\\\\ S_ (5) & \u003d & (\\ FRAC (137) (60)) & \\ APPROX & 2 (,) 283 \\ END (MATRIX))) | S 6 \u003d 49 20 \u003d 2, 45 S 7 \u003d 363 140 ≈ 2.593 S 8 \u003d 761 280 ≈ 2.718 S 10 3 ≈ 7.484 S 10 6 ≈ 14,393 (\\ DisplayStyle (\\ Begin (Matrix) S_ (6) & \u003d & ( \\ FRAC (49) (20)) & \u003d & 2 (,) 45 \\\\\\\\ S_ (7) & \u003d & (\\ FRAC (363) (140)) \\ Approx & 2 (,) 593 \\\\\\\\ S_ (8) & \u003d & (\\ FRAC (761) (280)) & \\ Approx & 2 (,) 718 \\\\\\\\ S_ (10 ^ (3)) \\ Approx & 7 (,) 484 \\\\\\\\ S_ ( 10 ^ (6)) & \\ Approx & 14 (,) 393 \\ END (MATRIX))) |
Formula Euler
With value ε n → 0 (\\ displaystyle \\ varepsilon _ (n) \\ rightarrow 0), therefore, for big N (\\ DisplayStyle N):
s n ≈ ln \u2061 (n) + γ (\\ displaystyle s_ (n) \\ approx \\ ln (n) + \\ gamma) - Euler formula for the first amount N (\\ DisplayStyle N) members of the harmonic series.| N (\\ DisplayStyle N) | S n \u003d Σ k \u003d 1 n 1 k (\\ displaystyle s_ (n) \u003d \\ sum _ (k \u003d 1) ^ (n) (\\ FRAC (1) (k))) | Ln \u2061 (n) + γ (\\ displaystyle \\ ln (n) + \\ gamma) | ε n (\\ displaystyle \\ varepsilon _ (n)), (%) |
| 10 | 2,93 | 2,88 | 1,7 |
| 25 | 3,82 | 3,80 | 0,5 |
More accurate asymptotic formula for partial sum of the harmonic series:
Sn ≍ ln \u2061 (n) + γ + 1 2 n - 1 12 n 2 + 1 120 N 4 - 1 252 N 6 ⋯ \u003d ln \u2061 (n) + γ + 1 2 n - σ k \u003d 1 ∞ b 2 k 2 kn 2 k (\\ displaystyle s_ (n) \\ asymp \\ ln (n) + \\ gamma + (\\ FRAC (1) (2N)) - (\\ FRAC (1) (12N ^ (2))) + (\\ \\ Sum _ (k \u003d 1) ^ (\\ infty) (\\ FRAC (B_ (2K)) (2K \\, n ^ (2k))))where B 2 K (\\ DisplayStyle B_ (2K)) - Numbers Bernoulli.This series is divided, however, the error of calculations on it never exceeds half of the first dumped member.
Theoretical numerical properties of partial sums
∀ N\u003e 1 S N ∉ N (\\ DisplayStyle \\ Forall N\u003e 1 \\; \\; \\; \\; s_ (n) \\ notin \\ mathbb (n))
Road divergence
S n → ∞ (\\ displaystyle s_ (n) \\ rightarrow \\ infty) for N → ∞ (\\ DisplayStyle N \\ Rightarrow \\ Infty)
Harmonic row diverges Very slow (in order for the partial amount exceeded 100, about 10,43 elements of the series are necessary).
The divergence of the harmonic series can be demonstrated by comparing it with telescopic near:
Vn \u003d ln \u2061 (n + 1) - ln \u2061 n \u003d ln \u2061 (1 + 1 n) ~ + ∞ 1 n (\\ displaystyle v_ (n) \u003d \\ ln (n + 1) - \\ ln n \u003d \\ ln \\ The partial sum of which is obviously equal to:,Σ i \u003d 1 n - 1 v i \u003d ln \u2061 n ~ s n (\\ displaystyle \\ sum _ (i \u003d 1) ^ (n-1) v_ (i) \u003d \\ ln n \\ Sim s_ (n))
Proof Oreman.The proof of divergence can be constructed by grouping terms as follows:
Σ k \u003d 1 ∞ 1 k \u003d 1 + [1 2] + [1 3 + 1 4] + [1 5 + 1 6 + 1 7 + 1 8] + [1 9 + ⋯] + ⋯\u003e 1 + [1 2] + [1 4 + 1 4] + [1 8 + 1 8 + 1 8 + 1 8] + [1 16 + ⋯] + ⋯ \u003d 1 + 1 2 + 1 2 + 1 2 + 1 2 + ⋯. (\\ DisplayStyle (\\ Begin (Aligned) \\ Sum _ (k \u003d 1) ^ (\\ infty) (\\ FRAC (1) (K)) & () \u003d 1+ \\ left [(\\ FRAC (1) (2) ) \\ Right] + \\ left [(\\ FRAC (1) (3)) + (\\ FRAC (1) (4)) \\ RIGHT] + \\ LEFT [(\\ FRAC (1) (5)) + (\\ FRAC (1) (6)) + (\\ FRAC (1) (7)) + (\\ FRAC (1) (8)) \\ RIGHT] + \\ LEFT [(\\ FRAC (1) (9)) + \\ CDOTS \\ (4)) \\ RIGHT] + \\ left [(\\ FRAC (1) (8)) + (\\ FRAC (1) (8)) + (\\ FRAC (1) (8)) + (\\ FRAC (1) (8)) \\ Right] + \\ left [(\\ FRAC (1) (16)) + \\ cdots \\ right] + \\ cdots \\\\ & () \u003d 1+ \\ (\\ FRAC (1) (2)) \\ ) (2)) \\ \\ quad + \\ \\ cdots. \\ END (Aligned)))
The last row is obviously diverged. This proof belongs to the medieval scientist Nicholas Ohole (approx. 1350).Alternative proof of divergence
We offer the reader to make sure that the mistakes of this evidence
difference between
-m harmonic number and natural logarithm N (\\ DisplayStyle N)It converges to the permanent Euler - Musteroni. N (\\ DisplayStyle N) The difference between various harmonic numbers is never equal to an integer and no harmonic number except
H 1 \u003d 1 (\\ DisplayStyle H_ (1) \u003d 1) is not integer.Related ranks
A number of Dirichle
Summarized harmonious side (or near Dirichle) call a number
Σ k \u003d 1 ∞ 1 k α \u003d 1 + 1 2 α + 1 3 α + 1 4 α + ⋯ + 1 k α + ⋯ (\\ displaystyle \\ sum _ (k \u003d 1) ^ (\\ infty) (\\ FRAC ( 1) (k ^ (\\ alpha))) \u003d 1 + (\\ FRAC (1) (2 ^ (\\ alpha))) + (\\ FRAC (1) (3 ^ (\\ alpha))) + (\\ FRAC ( 1) (4 ^ (\\ alpha))) + \\ CDOTS + (\\ FRAC (1) (k ^ (\\ alpha))) + \\ CDOTS).The generalized harmonic row diverges α ⩽ 1 (\\ displaystyle \\ alpha \\ leqslant 1) and convergence α\u003e 1 (\\ displaystyle \\ alpha\u003e 1) .
The sum of the generalized harmonic series of order α (\\ displayStyle \\ Alpha) equal to the value of the Riemann zeta function:
Σ k \u003d 1 ∞ 1 k α \u003d ζ (α) (\\ displaystyle \\ sum _ (k \u003d 1) ^ (\\ infty) (\\ FRAC (1) (k ^ (\\ alpha))) \u003d \\ Zeta (\\ alpha ))For even this value is clearly expressed in the number of pi, for example, ζ (2) \u003d π 2 6 (\\ displaystyle \\ zeta (2) \u003d (\\ FRAC (\\ pi ^ (2)) (6))), and already for α \u003d 3 its value is analytically unknown.
Another illustration of the displacement of the harmonic series can be the ratio ζ (1 + 1 n) ~ N (\\ displaystyle \\ zeta (1 + (\\ FRAC (1) (N))) \\ SIM N). Therefore, it is said that such a series has a probability of 1, and the sum of the row is a random value with interesting properties. For example, the probability density function calculated at +2 or -2 points is:
0,124 999 999 999 999 999 999 999 999 999 999 999 999 999 7 642 …,distinguishing from ⅛ on less than 10 -42.
"Omnounce" harmonic series
A number of Kempner (eng.)If we consider the harmonic row, in which only the terms are left, the denominators of which do not contain figures 9, it turns out that the remaining amount is to the number<80 . Более того, доказано, что если оставить слагаемые, не содержащие любой заранее выбранной последовательности цифр, то полученный ряд будет сходиться. Однако из этого будет ошибочно заключать о сходимости исходного гармонического ряда, так как с ростом разрядов в числе N (\\ DisplayStyle N), fewer terms is taken for the sum of the "thinned" row. That is, ultimately discarded the overwhelming majority of members forming the sum of the harmonic series so as not to exceed the geometric progression limiting on top.
Answer: A row diverges.
Example number 3.
Find the sum of the $ \\ sum \\ limits_ (n \u003d 1) ^ (\\ infty) \\ FRAC (2) ((2n + 1) (2n + 3)) $.
Since the lower limit of summation is 1, then the total member of the row is recorded under the sum of the amount: $ U_n \u003d \\ FRAC (2) ((2N + 1) (2N + 3)) $. Make a N-Mu partial sum of a number, i.e. We summarize the first $ n $ members of the specified numerical series:
$$ s_n \u003d u_1 + u_2 + u_3 + u_4 + \\ ldots + u_n \u003d \\ frac (2) (3 \\ CDOT 5) + \\ FRAC (2) (5 \\ CDOT 7) + \\ FRAC (2) (7 \\ CDOT 9 ) + \\ FRAC (2) (9 \\ CDOT 11) + \\ LDOTS + \\ FRAC (2) ((2N + 1) (2N + 3)). $$.
Why I write exactly $ \\ FRAC (2) (3 \\ CDOT 5) $, and not $ \\ FRAC (2) (15) $, will be clear from further narration. However, the recording of a partial amount either on iota did not bring us closer to the goal. We need to find $ \\ lim_ (n \\ to \\ infty) s_n $, but if we just write down:
$$ \\ LIM_ (n \\ to \\ infty) s_n \u003d \\ lim_ (n \\ to \\ infty) \\ left (\\ FRAC (2) (3 \\ CDOT 5) + \\ FRAC (2) (5 \\ CDOT 7) + \\ then this entry is completely true in shape, nothing will give us essentially. To find a limit, the expression of the partial amount must be simplified.
for this, there is a standard conversion consisting in decomposition of the fraction $ \\ FRAC (2) ((2n + 1) (2N + 3)) $, which represents a common member of the row, on elementary fractions. The issue of decomposition of rational fractions on elementary is devoted to a separate topic (see, for example, example number 3 on this page). Enclosing the fraction of $ \\ FRAC (2) ((2n + 1) (2n + 3)) $ per elementary fraction, we will have:
$$ \\ FRAC (2) ((2n + 1) (2N + 3)) \u003d \\ FRAC (A) (2N + 1) + \\ FRAC (B) (2N + 3) \u003d \\ FRAC (A \\ CDOT (2N +3) + B \\ Cdot (2n + 1)) ((2n + 1) (2N + 3)). $$.
We equate the sprumeren in the left and right parts of the equality obtained:
$$ 2 \u003d A \\ CDOT (2N + 3) + B \\ CDOT (2N + 1). $$.
To find the values \u200b\u200bof $ a $ and $ b $ there are two ways. You can reveal brackets and regroup the components, and you can simply substitute a certain suitable values \u200b\u200binstead of $ n $. For a variety in this example, let's go first, and as follows, we will substitute the private values \u200b\u200bof $ n $. Revealing brackets and rearrangement terms, we get:
$$ 2 \u003d 2AN + 3A + 2BN + B; \\\\ 2 \u003d (2a + 2b) n + 3a + b. $$.
In the left side of the equality in front of the $ n $ costs zero. If you want, the left part of equality for clarity can be represented as $ 0 \\ Cdot n + $ 2. Since the left part of equality in front of $ n $ is zero, and in the right part of the equity before $ n $ costs $ 2a + 2b $, we have the first equation: $ 2a + 2b \u003d 0 $. Immediately divide both parts of this equation by 2, having received $ a + b \u003d 0 $ after that.
Since in the left part of equality, the free member is 2, and in the right part of equality, the free member is equal to $ 3a + b $, then $ 3a + b \u003d $ 2. So, we have a system:
$$ \\ left \\ (\\ begin (aligned) & a + b \u003d 0; \\\\ & 3a + b \u003d 2. \\ End (Aligned) \\ Right. $$
Proof will be carried out by mathematical induction. In the first step, it is necessary to check whether the previous equality is made $ s_n \u003d \\ FRAC (1) (3) - \\ FRAC (1) (2N + 3) $ with $ n \u003d 1 $. We know that $ s_1 \u003d u_1 \u003d \\ FRAC (2) (15) $, but will the expression of $ \\ FRAC (1) (3) - \\ FRAC (1) (2N + 3) $ value $ \\ FRAC (2 ) (15) $, if we substitute $ n \u003d 1 $ into it? Check:
$$ \\ FRAC (1) (3) - \\ FRAC (1) (2N + 3) \u003d \\ FRAC (1) (3) - \\ FRAC (1) (2 \\ Cdot 1 + 3) \u003d \\ FRAC (1) (3) - \\ FRAC (1) (5) \u003d \\ FRAC (5-3) (15) \u003d \\ FRAC (2) (15). $$.
So, at $ n \u003d 1 $, the equality $ s_n \u003d \\ FRAC (1) (3) - \\ FRAC (1) (2N + 3) $ is made. This is the first step of the method of mathematical induction.
Suppose that at $ n \u003d k $, the equality is performed, i.e. $ S_k \u003d \\ FRAC (1) (3) - \\ FRAC (1) (2K + 3) $. We prove that this equality will be performed at $ n \u003d k + $ 1. To do this, consider $ s_ (k + 1) $:
$$ S_ (k + 1) \u003d s_k + u_ (k + 1). $$.
Since $ U_n \u003d \\ FRAC (1) (2N + 1) - \\ FRAC (1) (2N + 3) $, then $ U_ (K + 1) \u003d \\ FRAC (1) (2 (k + 1) + 1) - \\ FRAC (1) (2 (k + 1) +3) \u003d \\ FRAC (1) (2K + 3) - \\ FRAC (1) (2 (k + 1) +3) $. According to the above assumption of $ s_k \u003d \\ FRAC (1) (3) - \\ FRAC (1) (2K + 3) $, therefore, the formula $ s_ (k + 1) \u003d s_k + u_ (k + 1) $ will take the form:
$$ S_ (K + 1) \u003d S_K + U_ (K + 1) \u003d \\ FRAC (1) (3) - \\ FRAC (1) (2K + 3) + \\ FRAC (1) (2K + 3) - \\ $$.
Conclusion: Formula $ s_n \u003d \\ FRAC (1) (3) - \\ FRAC (1) (2N + 3) $ is true at $ n \u003d k + $ 1. Consequently, according to the method of mathematical induction, the formula $ s_n \u003d \\ FRAC (1) (3) - \\ FRAC (1) (2N + 3) $ is true for any $ n \\ inn $. Equality is proven.
In the standard course of higher mathematics, they are usually satisfied with the "trimming" of the declining terms, without requiring any evidence. So, we have received an expression for the N-th partial amount: $ s_n \u003d \\ FRAC (1) (3) - \\ FRAC (1) (2N + 3) $. Find $ \\ Lim_ (n \\ to \\ infty) s_n $:
Conclusion: The specified series converges and the sum of its $ s \u003d \\ FRAC (1) (3) $.
The second way to simplify the formula for the partial amount.
Honestly, I myself prefer this method :) Let's write a partial amount in the abbreviated version:
$$ s_n \u003d \\ Sum \\ Limits_ (k \u003d 1) ^ (n) u_k \u003d \\ sum \\ limits_ (k \u003d 1) ^ (n) \\ FRAC (2) ((2k + 1) (2k + 3)). $$.
We have earlier that $ U_k \u003d \\ FRAC (1) (2k + 1) - \\ FRAC (1) (2K + 3) $, so:
$$ s_n \u003d \\ sum \\ limits_ (k \u003d 1) ^ (n) \\ FRAC (2) ((2K + 1) (2K + 3)) \u003d \\ Sum \\ Limits_ (k \u003d 1) ^ (n) \\ left (\\ FRAC (1) (2K + 1) - \\ FRAC (1) (2K + 3) \\ Right). $$.
The amount of $ s_n $ contains a final number of terms, so we can rearrange them as we are pleased. I want to first fold all the terms of the type $ \\ FRAC (1) (2k + 1) $, and then then move to the term of the type $ \\ FRAC (1) (2k + 3) $. This means that the partial amount will be imagined in this form:
$$ S_N \u003d \\ FRAC (1) (3) - \\ FRAC (1) (5) + \\ FRAC (1) (5) - \\ FRAC (1) (7) + \\ FRAC (1) (7) - \\ \\\\ \u003d \\ FRAC (1) (3) + \\ FRAC (1) (5) + \\ FRAC (1) (7) + \\ FRAC (1) (9) + \\ LDOTS + \\ FRAC (1) (2N + 1 ) - \\ left (\\ FRAC (1) (5) + \\ FRAC (1) (7) + \\ FRAC (1) (9) + \\ LDOTS + \\ FRAC (1) (2N + 3) \\ Right). $$.
Of course, the deployed record is extremely uncomfortable, therefore the equality presented above can be issued more compact:
$$ S_n \u003d \\ Sum \\ Limits_ (k \u003d 1) ^ (n) \\ left (\\ FRAC (1) (2K + 1) - \\ FRAC (1) (2K + 3) \\ Right) \u003d \\ Sum \\ Limits_ ( k \u003d 1) ^ (n) \\ FRAC (1) (2K + 1) - \\ Sum \\ Limits_ (k \u003d 1) ^ (n) \\ FRAC (1) (2K + 3). $$.
Now we transform the expressions $ \\ FRAC (1) (2K + 1) $ and $ \\ FRAC (1) (2K + 3) $ to one appearance. I suppose comfortable to lead to the form of a larger fraction (although it is possible to smallest, this is a matter of taste). Since $ \\ FRAC (1) (2K + 1)\u003e \\ FRAC (1) (2K + 3) $ (the greater the denominator, the smaller the fraction), then we will drive $ \\ FRAC (1) (2k + 3) $ to the type $ \\ FRAC (1) (2k + 1) $.
Expression in the denomoter denutor $ \\ FRAC (1) (2K + 3) $ I will present in this form:
$$ \\ FRAC (1) (2K + 3) \u003d \\ FRAC (1) (2K + 2 + 1) \u003d \\ FRAC (1) (2 (k + 1) +1). $$.
And the amount of $ \\ sum \\ limits_ (k \u003d 1) ^ (n) \\ FRAC (1) (2K + 3) $ can now be written as:
$$ \\ Sum \\ Limits_ (k \u003d 1) ^ (n) \\ FRAC (1) (2K + 3) \u003d \\ Sum \\ Limits_ (k \u003d 1) ^ (n) \\ FRAC (1) (2 (k + 1 ) +1) \u003d \\ Sum \\ Limits_ (k \u003d 2) ^ (n + 1) \\ FRAC (1) (2K + 1). $$.
If the equality is $ \\ sum \\ limits_ (k \u003d 1) ^ (n) \\ FRAC (1) (2K + 3) \u003d \\ Sum \\ Limits_ (k \u003d 2) ^ (n + 1) \\ FRAC (1) (2k + 1) $ does not cause questions, then let's go further. If you have questions, I ask you to deploy a note.
How did we get the transformed amount? Show \\ Hide
We had a number of $ \\ sum \\ limits_ (k \u003d 1) ^ (n) \\ FRAC (1) (2k + 3) \u003d \\ Sum \\ Limits_ (k \u003d 1) ^ (n) \\ FRAC (1) (2 (2 k + 1) +1) $. Let's introduce a new variable instead of $ k + 1 $, for example, $ T $. So, $ t \u003d k + $ 1.
How was the old variable $ k $ changed? And it changed from 1 to $ n $. Let's find out how the new variable $ t will change. If $ k \u003d 1 $, then $ t \u003d 1 + 1 \u003d $ 2. If $ k \u003d n $, then $ t \u003d n + 1 $. So, the expression $ \\ sum \\ limits_ (k \u003d 1) ^ (n) \\ FRAC (1) (2 (k + 1) +1) $ now it was: $ \\ sum \\ limits_ (t \u003d 2) ^ (n +1) \\ FRAC (1) (2t + 1) $.
$$ \\ SUM \\ Limits_ (k \u003d 1) ^ (n) \\ FRAC (1) (2 (k + 1) +1) \u003d \\ Sum \\ Limits_ (T \u003d 2) ^ (n + 1) \\ FRAC (1 ) (2T + 1). $$.
We have the amount of $ \\ sum \\ limits_ (t \u003d 2) ^ (n + 1) \\ FRAC (1) (2t + 1) $. Question: Isn't it equal to how to use in this sum? :) Trolly recording the letter $ k $ instead of $ t $, we get the following:
$$ \\ SUM \\ LIMITS_ (T \u003d 2) ^ (n + 1) \\ FRAC (1) (2T + 1) \u003d \\ Sum \\ Limits_ (k \u003d 2) ^ (n + 1) \\ FRAC (1) (2K +1). $$.
So it turns out the equality of $ \\ sum \\ limits_ (k \u003d 1) ^ (n) \\ FRAC (1) (2 (k + 1) +1) \u003d \\ Sum \\ Limits_ (k \u003d 2) ^ (n + 1) \\ FRAC (1) (2K + 1) $.
Thus, a partial amount can be represented as follows:
$$ S_n \u003d \\ Sum \\ Limits_ (k \u003d 1) ^ (n) \\ FRAC (1) (2K + 1) - \\ Sum \\ Limits_ (k \u003d 1) ^ (n) \\ FRAC (1) (2K + 3 ) \u003d \\ Sum \\ Limits_ (k \u003d 1) ^ (n) \\ FRAC (1) (2K + 1) - \\ Sum \\ Limits_ (k \u003d 2) ^ (n + 1) \\ FRAC (1) (2K + 1 ). $$.
Note that the amount of $ \\ sum \\ limits_ (k \u003d 1) ^ (n) \\ FRAC (1) (2K + 1) $ and $ \\ sum \\ limits_ (k \u003d 2) ^ (n + 1) \\ FRAC (1 ) (2K + 1) $ is different only within the summation. Let's make these limits are the same. "Taking" the first element from the amount of $ \\ sum \\ limits_ (k \u003d 1) ^ (n) \\ FRAC (1) (2k + 1) $ will have:
$$ \\ SUM \\ LIMITS_ (k \u003d 1) ^ (n) \\ FRAC (1) (2K + 1) \u003d \\ FRAC (1) (2 \\ CDot 1 + 1) + \\ Sum \\ Limits_ (k \u003d 2) ^ (n) \\ FRAC (1) (2K + 1) \u003d \\ FRAC (1) (3) + \\ Sum \\ Limits_ (k \u003d 2) ^ (n) \\ FRAC (1) (2K + 1). $$.
"Taking" the last element from the amount of $ \\ Sum \\ Limits_ (k \u003d 2) ^ (n + 1) \\ FRAC (1) (2K + 1) $, we get:
$$ \\ Sum \\ Limits_ (k \u003d 2) ^ (n + 1) \\ FRAC (1) (2K + 1) \u003d \\ Sum \\ Limits_ (k \u003d 2) ^ (n) \\ FRAC (1) (2K + 1 ) + \\ FRAC (1) (2 (n + 1) +1) \u003d \\ Sum \\ Limits_ (k \u003d 2) ^ (n) \\ FRAC (1) (2K + 1) + \\ FRAC (1) (2N + 3). $$
Then the expression for the partial amount will take the form:
$$ S_n \u003d \\ Sum \\ Limits_ (k \u003d 1) ^ (n) \\ FRAC (1) (2K + 1) - \\ Sum \\ Limits_ (k \u003d 2) ^ (n + 1) \\ FRAC (1) (2K +1) \u003d \\ FRAC (1) (3) + \\ Sum \\ Limits_ (k \u003d 2) ^ (n) \\ FRAC (1) (2k + 1) - \\ left (\\ sum \\ limits_ (k \u003d 2) ^ (n) \\ FRAC (1) (2K + 1) + \\ FRAC (1) (2N + 3) \\ RIGHT) \u003d \\\\ \u003d \\ FRAC (1) (3) + \\ Sum \\ Limits_ (k \u003d 2) ^ (n) \\ FRAC (1) (2K + 1) - \\ Sum \\ Limits_ (k \u003d 2) ^ (n) \\ FRAC (1) (2K + 1) - \\ FRAC (1) (2N + 3) \u003d \\ $$.
If you miss all the explanations, the process of finding the abbreviated formula for the N-th partial amount will take this kind:
$$ s_n \u003d \\ Sum \\ Limits_ (k \u003d 1) ^ (n) u_k \u003d \\ sum \\ limits_ (k \u003d 1) ^ (n) \\ FRAC (2) ((2K + 1) (2K + 3)) \u003d \\ Sum \\ Limits_ (k \u003d 1) ^ (n) \\ left (\\ FRAC (1) (2K + 1) - \\ FRAC (1) (2K + 3) \\ Right) \u003d \\\\ \u003d \\ SUM \\ LIMITS_ (K \u003d 1) ^ (n) \\ FRAC (1) (2K + 1) - \\ Sum \\ Limits_ (k \u003d 1) ^ (n) \\ FRAC (1) (2K + 3) \u003d \\ FRAC (1) (3) + \\ Sum \\ Limits_ (k \u003d 2) ^ (n) \\ FRAC (1) (2K + 1) - \\ left (\\ sum \\ limits_ (k \u003d 2) ^ (n) \\ FRAC (1) (2K + 1 ) + \\ FRAC (1) (2N + 3) \\ RIGHT) \u003d \\ FRAC (1) (3) - \\ FRAC (1) (2N + 3). $$.
Let me remind you that we crossed $ \\ FRAC (1) (2k + 3) $ to the type $ \\ FRAC (1) (2K + 1) $. Of course, you can proceed on the contrary, i.e. Present the fraction of $ \\ FRAC (1) (2K + 1) $ in the form $ \\ FRAC (1) (2K + 3) $. The final expression for the partial amount will not change. The process of finding a partial amount in this case I will hide under Note.
How to find $ s_n $ if you give the mind of another fraction? Show \\ Hide
$$ S_n \u003d \\ Sum \\ Limits_ (k \u003d 1) ^ (n) \\ FRAC (1) (2K + 1) - \\ Sum \\ Limits_ (k \u003d 1) ^ (n) \\ FRAC (1) (2K + 3 ) \u003d \\ Sum \\ Limits_ (k \u003d 0) ^ (n - 1) \\ FRAC (1) (2K + 3) - \\ Sum \\ Limits_ (k \u003d 1) ^ (n) \\ FRAC (1) (2K + 3 ) \u003d \\\\ \u003d \\ FRAC (1) (3) + \\ sum \\ limits_ (k \u003d 1) ^ (n - 1) \\ FRAC (1) (2K + 3) - \\ left (\\ Sum \\ Limits_ (k \u003d 1) ^ (n - 1) \\ FRAC (1) (2K + 3) + \\ FRAC (1) (2N + 3) \\ Right) \u003d \\ FRAC (1) (3) - \\ FRAC (1) (2N + 3). $$.
So, $ s_n \u003d \\ FRAC (1) (3) - \\ FRAC (1) (2N + 3) $. We find the limit of $ \\ lim_ (n \\ to \\ infty) s_n $:
$$ \\ LIM_ (n \\ to \\ infty) s_n \u003d \\ lim_ (n \\ to \\ infty) \\ left (\\ FRAC (1) (3) - \\ FRAC (1) (2N + 3) \\ Right) \u003d \\ FRAC (1) (3) -0 \u003d \\ FRAC (1) (3). $$.
The specified series converges and the sum of its $ s \u003d \\ FRAC (1) (3) $.
Answer: $ S \u003d \\ FRAC (1) (3) $.
The continuation of the topic of finding the amount of the number will be considered in the second and third parts.
This article is a structured and detailed information that can be useful during the analysis of exercises and tasks. We will consider the topic of numerical rows.
This article begins with basic definitions and concepts. Next, we are standard options and study the basic formulas. In order to secure the material, the article presents the main examples and tasks.
Basic theses
To begin with, imagine the system: A 1, A 2. . . , a n ,. . . , where a k ∈ R, k \u003d 1, 2. . . .
For example, we take such numbers as: 6, 3, - 3 2, 3 4, 3 8, - 3 16 ,. . . .
Definition 1.
Numeric row is the sum of members σ a k k \u003d 1 ∞ \u003d a 1 + a 2 +. . . + a n +. . . .
To better understand the definition, consider this case in which Q \u003d - 0. 5: 8 - 4 + 2 - 1 + 1 2 - 1 4 +. . . \u003d Σ k \u003d 1 ∞ (- 16) · - 1 2 k.
Definition 2.
a K is common or k - Member of a number.
It looks about in this way - 16 · 1 2 K.
Definition 3.
Partial sum of the row It looks roughly thus s n \u003d a 1 + a 2 +. . . + a n in which N. - Love number. S n is N -O. sum of a number.
For example, Σ k \u003d 1 ∞ (- 16) · - 1 2 K is s 4 \u003d 8 - 4 + 2 - 1 \u003d 5.
S 1, S 2 ,. . . , S n ,. . . Forming an infinite sequence of a numeric row.
For row N-having The amount is located according to the formula S n \u003d A 1 · (1 - Q n) 1 - Q \u003d 8 · 1 - - 1 2 N 1 - - 1 2 \u003d 16 3 · 1 - - 1 2 n. Use the following sequence of partial sums: 8, 4, 6, 5 ,. . . , 16 3 · 1 - - 1 2 N ,. . . .
Definition 4.
A series Σ k \u003d 1 ∞ a k is convergent Then when the sequence has the final limit S \u003d Lim S N n → + ∞. If there is no limit or the sequence of infinite, then the range Σ k \u003d 1 ∞ A K is called drawn.
Definition 5.
The sum of the converging row Σ k \u003d 1 ∞ A k is the limit of the sequence σ k \u003d 1 ∞ a k \u003d lim s n n → + ∞ \u003d s.
In this example, Lim S nn → + ∞ \u003d Lim 16 3 t → + ∞ · 1 - 1 2 n \u003d 16 3 · Lim n → + ∞ 1 - - 1 2 n \u003d 16 3, a number σ k \u003d 1 ∞ (- 16) · - 1 2 K converge. The amount is 16 3: σ k \u003d 1 ∞ (- 16) · - 1 2 k \u003d 16 3.
Example 1.
As an example, the diverging series can be given a sum of geometric progression with a denominator greater than a unit: 1 + 2 + 4 + 8 +. . . + 2 N - 1 +. . . \u003d Σ k \u003d 1 ∞ 2 k - 1.
n-having partial amount is determined by the expression S n \u003d a 1 · (1 - Qn) 1 - Q \u003d 1 · (1 - 2 n) 1 - 2 \u003d 2 n - 1, and the limit of partial sums is infinite: Lim n → + ∞ S n \u003d Lim n → + ∞ (2 n - 1) \u003d + ∞.
Another example of a diverging numerical series is the sum of the form Σ k \u003d 1 ∞ 5 \u003d 5 + 5 +. . . . In this case, the n-way partial amount can be calculated as s n \u003d 5 n. The limit of partial sums is infinite with Lim N → + ∞ S n \u003d Lim n → + ∞ 5 n \u003d + ∞.
Definition 6.
The sum of this species as σ k \u003d 1 ∞ \u003d 1 + 1 2 + 1 3 +. . . + 1 N +. . . - this is harmonic numeric row.
Definition 7.
The sum Σ k \u003d 1 ∞ 1 k s \u003d 1 + 1 2 s + 1 3 s +. . . + 1 N S +. . . where S. - Active number is generalized by harmonious numeric.
Definitions discussed above will help you to solve most examples and tasks.
In order to add definitions, it is necessary to prove certain equations.
- Σ k \u003d 1 ∞ 1 K - consigned.
We act by the method from the opposite. If it converges, the limit is finite. You can write the equation as lim n → + ∞ s n \u003d s and lim n → + ∞ s 2 n \u003d s. After certain actions, we obtain the equality l i m n → + ∞ (S 2 n - s n) \u003d 0.
On the contrary
S 2 n - S n \u003d 1 + 1 2 + 1 3 +. . . + 1 n + 1 n + 1 + 1 n + 2 +. . . + 1 2 N - - 1 + 1 2 + 1 3 +. . . + 1 n \u003d 1 n + 1 + 1 n + 2 +. . . + 1 2 n
Fair the following inequalities 1 n + 1\u003e 1 2 N, 1 n + 1\u003e 1 2 N ,. . . , 1 2 n - 1\u003e 1 2 n. We obtain that S 2 n - S n \u003d 1 n + 1 + 1 n + 2 +. . . + 1 2 N\u003e 1 2 N + 1 2 N +. . . + 1 2 n \u003d n 2 n \u003d 1 2. The expression S 2 n - S n\u003e 1 2 indicates that Lim n → + ∞ (S 2 n - S n) \u003d 0 is not achieved. Row consigned.
- b 1 + B 1 Q + B 1 Q 2 +. . . + B 1 Q n +. . . \u003d Σ k \u003d 1 ∞ b 1 q k - 1
It is necessary to confirm that the sum of the sequence of numbers converges when Q< 1 , и расходится при q ≥ 1 .
According to the above definitions, the amount n. Members are determined according to the formula S n \u003d B 1 · (Q n - 1) Q - 1.
If Q.< 1 верно
lim N → + ∞ S n \u003d Lim N → + ∞ B 1 · Qn - 1 Q - 1 \u003d B 1 · Lim N → + ∞ Qnq - 1 - Lim N → + ∞ 1 Q - 1 \u003d B 1 · 0 - 1 Q - 1 \u003d B 1 Q - 1
We have proven that the numerical series converges.
At q \u003d 1 b 1 + b 1 + b 1 +. . . Σ k \u003d 1 ∞ b 1. Amounts can be found using the formula S n \u003d B 1 · n, the limit is infinite with Lim N → + ∞ S n \u003d Lim n → + ∞ B 1 · n \u003d ∞. In the presented version, a number diverged.
If a Q \u003d - 1, then the row looks like B 1 - B 1 + B 1 -. . . \u003d Σ k \u003d 1 ∞ B 1 (- 1) k + 1. Partial sums look like S n \u003d B 1 for odd N., and S n \u003d 0 for even N.. Having considered this case, we will make sure that the limit is not and the number is divergent.
With q\u003e 1, Lim N → + ∞ S N \u003d Lim N → + ∞ B 1 · (Qn - 1) Q - 1 \u003d B 1 · Lim N → + ∞ QNQ - 1 - Lim N → + ∞ 1 Q - 1 \u003d B 1 · ∞ - 1 Q - 1 \u003d ∞
We have proven that the numerical row is diverged.
- A series Σ k \u003d 1 ∞ 1 k s converges if S\u003e 1. And diverges if S ≤ 1.
For S \u003d 1. We obtain σ k \u003d 1 ∞ 1 k, the row diverges.
At S.< 1 получаем 1 k s ≥ 1 k для k,natural number. Since the series is divergent σ k \u003d 1 ∞ 1 k, then the limit is not. Following this, the sequence σ k \u003d 1 ∞ 1 k s is unlimited. We conclude that the selected row diverges when S.< 1 .
It is necessary to provide evidence that the series Σ k \u003d 1 ∞ 1 k s converges when S\u003e 1..
Imagine S 2 N - 1 - S N - 1:
S 2 N - 1 - S n - 1 \u003d 1 + 1 2 S + 1 3 S +. . . + 1 (n - 1) s + 1 n s + 1 (n + 1) s +. . . + 1 (2 n - 1) S - - 1 + 1 2 S + 1 3 S +. . . + 1 (n - 1) s \u003d 1 n s + 1 (n + 1) s +. . . + 1 (2 n - 1) s
Suppose that 1 (n + 1) s< 1 n s , 1 (n + 2) s < 1 n s , . . . , 1 (2 n - 1) s < 1 n s , тогда S 2 n - 1 - S n - 1 = 1 n s + 1 (n + 1) s + . . . + 1 (2 n - 1) s < < 1 n s + 1 n s + . . . + 1 n s = n n s = 1 n s - 1
Represent an equation for numbers that are natural and even n \u003d 2: s 2 n - 1 - s n - 1 \u003d s 3 - s 1 \u003d 1 2 s + 1 3 s< 1 2 s - 1 n = 4: S 2 n - 1 - S n - 1 = S 7 - S 3 = 1 4 s + 1 5 s + 1 6 s + 1 7 s < 1 4 s - 1 = 1 2 s - 1 2 n = 8: S 2 n - 1 - S n - 1 = S 15 - S 7 = 1 8 s + 1 9 s + . . . + 1 15 s < 1 8 s - 1 = 1 2 s - 1 3 . . .
We get:
Σ k \u003d 1 ∞ 1 k s \u003d 1 + 1 2 s + 1 3 s + 1 4 s +. . . + 1 7 S + 1 8 S +. . . + 1 15 S +. . . \u003d \u003d 1 + S 3 - s 1 + S 7 - S 3 + S 15 + S 7 +. . .< < 1 + 1 2 s - 1 + 1 2 s - 1 2 + 1 2 s - 1 3 + . . .
Expression 1 + 1 2 S - 1 + 1 2 S - 1 2 + 1 2 S - 1 3 +. . . - This is the sum of geometric progression Q \u003d 1 2 S - 1. According to the initial data when S\u003e 1., then 0< q < 1 . Получаем, ∑ k = 1 ∞ < 1 + 1 2 s - 1 + 1 2 s - 1 2 + 1 2 s - 1 3 + . . . = 1 1 - q = 1 1 - 1 2 s - 1 . Последовательность ряда при S\u003e 1. It increases and limited from above 1 1 - 1 2 S - 1. Imagine that there is a limit and a number is a converging σ k \u003d 1 ∞ 1 K s.
Definition 8.
Rods Σ k \u003d 1 ∞ a k aligning in the eventif its members\u003e 0 a k\u003e 0, k \u003d 1, 2 ,. . . .
Rods Σ k \u003d 1 ∞ b k AligningIf the signs of numbers differ. This example is represented as Σ k \u003d 1 ∞ bk \u003d Σ k \u003d 1 ∞ (- 1) k · ak or σ k \u003d 1 ∞ bk \u003d σ k \u003d 1 ∞ (- 1) k + 1 · ak, where AK\u003e 0 , k \u003d 1, 2 ,. . . .
Rods Σ k \u003d 1 ∞ b k SignSince it has many numbers, negative and positive.
The second version of the row is a third option case.
We give examples for each case, respectively:
6 + 3 + 3 2 + 3 4 + 3 8 + 3 16 + . . . 6 - 3 + 3 2 - 3 4 + 3 8 - 3 16 + . . . 6 + 3 - 3 2 + 3 4 + 3 8 - 3 16 + . . .
For the third version, absolute and conditional convergence can also be determined.
Definition 9.
The alternating series Σ k \u003d 1 ∞ b k is absolutely converged in the case when σ k \u003d 1 ∞ b K is also considered to be convergent.
We will analyze several characteristic options in detail.
Example 2.
If rows 6 - 3 + 3 2 - 3 4 + 3 8 - 3 16 +. . . and 6 + 3 - 3 2 + 3 4 + 3 8 - 3 16 +. . . Defined as converging, it is correct to assume that 6 + 3 + 3 2 + 3 4 + 3 8 + 3 16 +. . .
Definition 10.
The alternating series Σ k \u003d 1 ∞ b K is considered conditionally convergent if σ k \u003d 1 ∞ b is consigned, and the range Σ k \u003d 1 ∞ b K is considered to be converging.
Example 3.
In detail we analyze the variant σ k \u003d 1 ∞ (- 1) k + 1 k \u003d 1 - 1 2 + 1 3 - 1 4 +. . . . A series Σ k \u003d 1 ∞ (- 1) k + 1 k \u003d σ k \u003d 1 ∞ 1 k, which consists of absolute values, is defined as consider. This option is considered to be converging, as it is easy to determine. From this example, we learn that the series Σ k \u003d 1 ∞ (- 1) k + 1 k \u003d 1 - 1 2 + 1 3 - 1 4 +. . . It will be considered conditionally convergent.
Features of converging series
We analyze properties for certain cases
- If Σ k \u003d 1 ∞ a k will be converged, then the series Σ k \u003d m + 1 ∞ a k is also recognized as convergent. It can be noted that a row without M. Members are also considered to be convergent. In case we add to σ k \u003d m + 1 ∞ a k several numbers, the resulting result will also be converging.
- If Σ k \u003d 1 ∞ a k converges and sum \u003d S., then a series of Σ k \u003d 1 ∞ a · a k, σ k \u003d 1 ∞ a · a k \u003d a · s, where A. -constant.
- If σ k \u003d 1 ∞ a k and σ k \u003d 1 ∞ b k are convergent, amounts A. and B.also, then the rows σ k \u003d 1 ∞ a k + b k and σ k \u003d 1 ∞ a k - b K are also converge. Amounts will be equal A + B. and A - B. respectively.
Determine that the series converges σ k \u003d 1 ∞ 2 3 k · k 3.
Change the expression Σ k \u003d 1 ∞ 2 3 k · k 3 \u003d σ k \u003d 1 ∞ 2 3 · 1 K 4 3. A series Σ k \u003d 1 ∞ 1 k 4 3 is considered to be converging, since the series Σ k \u003d 1 ∞ 1 k s converges when S\u003e 1.. In accordance with the second property, Σ k \u003d 1 ∞ 2 3 · 1 K 4 3.
Example 5.
Determine whether the series Σ n \u003d 1 ∞ 3 + N n 5 2 converges.
We transform the original option Σ n \u003d 1 ∞ 3 + N n 5 2 \u003d Σ n \u003d 1 ∞ 3 n 5 2 + N n 2 \u003d Σ n \u003d 1 ∞ 3 N 5 2 + Σ n \u003d 1 ∞ 1 n 2.
We obtain the sum Σ n \u003d 1 ∞ 3 N 5 2 and Σ n \u003d 1 ∞ 1 N 2. Each series is recognized as convergent according to the property. So, as the ranks agree, the initial version too.
Example 6.
Calculate, whether the series 1 is 6 + 1 2 - 2 + 1 4 - 2 3 + 1 8 - 2 9 +. . . and calculate the amount.
Specify the original option:
1 - 6 + 1 2 - 2 + 1 4 - 2 3 + 1 8 - 2 9 +. . . \u003d 1 + 1 2 + 1 4 + 1 8 +. . . - 2 · 3 + 1 + 1 3 + 1 9 +. . . \u003d \u003d Σ k \u003d 1 ∞ 1 2 k - 1 - 2 · σ k \u003d 1 ∞ 1 3 k - 2
Each series converges, as it is one of the members of the numerical sequence. According to the third property, we can calculate that the initial version is also converging. We calculate the amount: the first term of the row σ k \u003d 1 ∞ 1 2 k - 1 \u003d 1, and the denominator \u003d 0. 5, followed by, Σ k \u003d 1 ∞ 1 2 k - 1 \u003d 1 1 - 0. 5 \u003d 2. The first term σ k \u003d 1 ∞ 1 3 k - 2 \u003d 3, and the denominator of the decreasing numerical sequence \u003d 1 3. We obtain: Σ k \u003d 1 ∞ 1 3 k - 2 \u003d 3 1 - 1 3 \u003d 9 2.
Use the expressions obtained above in order to determine the amount 1 - 6 + 1 2 - 2 + 1 4 - 2 3 + 1 8 - 2 9 +. . . \u003d Σ k \u003d 1 ∞ 1 2 k - 1 - 2 · σ k \u003d 1 ∞ 1 3 k - 2 \u003d 2 - 2 · 9 2 \u003d - 7
The necessary condition for determining is a series of convergent
Definition 11.If the series Σ k \u003d 1 ∞ a k is convergent, then its limit k-mo Member \u003d 0: Lim k → + ∞ a k \u003d 0.
If we check any option, you need to not forget about the indispensable condition. If it is not executed, the row diverges. If Lim K → + ∞ A k ≠ 0, then the row is consigned.
It should be clarified that the condition is important, but not enough. If the equality Lim k → + ∞ a k \u003d 0 is performed, this does not ensure that σ k \u003d 1 ∞ a k is convergent.
Let us give an example. For the harmonic series Σ k \u003d 1 ∞ 1 K, the condition is performed by Lim K → + ∞ 1 k \u003d 0, but the row is still diverted.
Example 7.
Determine the convergence of σ n \u003d 1 ∞ n 2 1 + n.
Check the initial expression on the implementation of the condition Lim n → + ∞ N 2 1 + n \u003d Lim n → + ∞ n 2 n 2 1 n 2 + 1 n \u003d lim n → + ∞ 1 1 n 2 + 1 n \u003d 1 + 0 + 0 \u003d + ∞ ≠ 0
Limit n-one Member is not equal to 0. We have proven that this series diverges.
How to determine the convergence of the alignmental series.
If you constantly use the specified signs, you will have to constantly calculate the limits. This section will help avoid difficulties while solving examples and tasks. In order to determine the convergence of the alignmental series, there is a specific condition.
For convergence of the alignment, σ k \u003d 1 ∞ a k, a k\u003e 0 ∀ k \u003d 1, 2, 3 ,. . . It is necessary to determine the limited sequence of amounts.
How to compare rows
There are several signs of comparing rows. We compare a number, the convergence of which is proposed to determine, with that close, the convergence of which is known.
First feature
Σ k \u003d 1 ∞ a k and σ k \u003d 1 ∞ b k - aligning rows. Inequality a k ≤ b k is valid for k \u003d 1, 2, 3, ... From this it follows that from the row σ k \u003d 1 ∞ b k we can obtain σ k \u003d 1 ∞ a k. Since Σ k \u003d 1 ∞ A k is dispelled, then the range σ k \u003d 1 ∞ b k can be defined as consider.
This rule is constantly used to solve equations and is a serious argument that will help determine convergence. Difficulties may be to choose the appropriate example for comparison can be found far from every case. Quite often, the number is selected according to the principle according to which the indicator k-mo A member will be equal to the result of subtracting the indicators of the degree of the numerator and the denominator k-mo Member of a series. Suppose that a k \u003d k 2 + 3 4 k 2 + 5, the difference will be equal to 2 – 3 = - 1 . In this case, it can be determined that for comparison, a number are needed with k - A member B k \u003d k - 1 \u003d 1 k, which is harmonious.
In order to secure the material obtained, consider in detail a pair of typical options.
Example 8.
Determine what is a series Σ k \u003d 1 ∞ 1 k - 1 2.
Since the limit \u003d 0 Lim k → + ∞ 1 k - 1 2 \u003d 0, we performed the necessary condition. Inequality will be fair 1 K< 1 k - 1 2 для k,which are natural. From the preceding paragraphs, we learned that the harmonic series Σ k \u003d 1 ∞ 1 K is consigned. According to the first character, it can be proved that the initial version is divergent.
Example 9.
Determine is a series of convergent or diverging Σ k \u003d 1 ∞ 1 K 3 + 3 K - 1.
In this example, a necessary condition is performed, since Lim K → + ∞ 1 K 3 + 3 K - 1 \u003d 0. Introducing 1 K 3 + 3 K - 1 in the form of inequality< 1 k 3 для любого значения K.. A series Σ k \u003d 1 ∞ 1 K 3 is convergent, since the harmonic series Σ k \u003d 1 ∞ 1 k s converges when S\u003e 1.. According to the first sign, we can conclude that the numerical series is convergent.
Example 10.
Determine, is what a number σ k \u003d 3 ∞ 1 k ln (Ln k) is. Lim K → + ∞ 1 k ln (Ln k) \u003d 1 + ∞ + ∞ \u003d 0.
In this embodiment, you can mark the implementation of the desired condition. We define a row for comparison. For example, Σ k \u003d 1 ∞ 1 K s. To determine what is equal to the degree, we will list the sequence (Ln (Ln k)), k \u003d 3, 4, 5. . . . Members of the sequence ln (ln 3), ln (ln 4), ln (ln 5) ,. . . Increases indefinitely. After analyzing the equation, it can be noted that by taking N \u003d 1619 as a value, then members of the sequence\u003e 2. For this sequence, the inequality 1 K ln (LN K) will be true< 1 k 2 . Ряд ∑ k = N ∞ 1 k 2 сходится согласно первому признаку, так как ряд ∑ k = 1 ∞ 1 k 2 тоже сходящийся. Отметим, что согласно первому признаку ряд ∑ k = N ∞ 1 k ln (ln k) сходящийся. Можно сделать вывод, что ряд ∑ k = 3 ∞ 1 k ln (ln k) также сходящийся.
Second sign
Suppose that Σ k \u003d 1 ∞ a k and σ k \u003d 1 ∞ b k are aligning numerical rows.
If lim k → + ∞ a k b k ≠ ∞, then the range Σ k \u003d 1 ∞ b k converges, and σ k \u003d 1 ∞ a k converges as well.
If lim k → + ∞ a k b k ≠ 0, since the series Σ k \u003d 1 ∞ b k is dispelled, then σ k \u003d 1 ∞ a k is also diverted.
If Lim K → + ∞ A k b k ≠ ∞ and lim k → + ∞ a k b ≠ 0, then the convergence or divergence of the row means the convergence or divergence of the other.
Consider Σ k \u003d 1 ∞ 1 k 3 + 3 k - 1 using the second feature. For comparison σ k \u003d 1 ∞ b k, we take a row of Σ k \u003d 1 ∞ 1 k 3. We define the limit: Lim k → + ∞ a k b k \u003d lim k → + ∞ 1 k 3 + 3 k - 1 1 k 3 \u003d lim k → + ∞ k 3 k 3 + 3 k - 1 \u003d 1
According to the second feature, it can be determined that a series of σ k \u003d 1 ∞ 1 K 3 is indicated that the initial option also converges.
Example 11.
Determine what is a series Σ n \u003d 1 ∞ k 2 + 3 4 K 3 + 5.
We analyze the necessary condition Lim K → ∞ k 2 + 3 4 K 3 + 5 \u003d 0, which is performed in this embodiment. According to the second feature, we take a number σ k \u003d 1 ∞ 1 k. We are looking for a limit: Lim k → + ∞ k 2 + 3 4 K 3 + 5 1 k \u003d lim k → + ∞ k 3 + 3 k 4 k 3 + 5 \u003d 1 4
According to the above theses, the consistent row entails the divergence of the original series.
Third sign
Consider the third sign of comparison.
Suppose that σ k \u003d 1 ∞ a k and _ σ k \u003d 1 ∞ b k is the alignmental numeric rows. If the condition is performed for a certain number a k + 1 a k ≤ b k + 1 b k, then the convergence of this series Σ k \u003d 1 ∞ b k means that the series Σ k \u003d 1 ∞ A K is also converging. The consisting of a series Σ k \u003d 1 ∞ a k entails the divergence of σ k \u003d 1 ∞ b k.
Sign of Dalamber
Imagine that σ k \u003d 1 ∞ a k is a sign-plating numerical series. If Lim K → + ∞ A k + 1 A K< 1 , то ряд является сходящимся, если lim k → + ∞ a k + 1 a k > 1, then divergent.
Note 1.
The sign of the Dalamber is valid if the limit is infinite.
If lim k → + ∞ a k + 1 a k \u003d - ∞, then the series is converging if Lim K → ∞ A k + 1 a k \u003d + ∞, then divergent.
If lim k → + ∞ a k + 1 a k \u003d 1, the sign of the Dalamber will not help and need to spend some more studies.
Example 12.
Determine, is a series of converging or divergent σ k \u003d 1 ∞ 2 k + 1 2 K on the basis of the Dalamber.
It is necessary to check whether the necessary condition for convergence is performed. We calculate the limit by using the Lopital Rule: Lim K → + ∞ 2 k + 1 2 k \u003d ∞ ∞ \u003d Lim k → + ∞ 2 k + 1 "2 k" \u003d Lim K → + ∞ 2 2 k · ln 2 \u003d 2 + ∞ · ln 2 \u003d 0
We can see that the condition is performed. We use the sign of the Dalamber: Lim K → + ∞ \u003d Lim K → + ∞ 2 (k + 1) + 1 2 k + 1 2 k + 1 2 k \u003d 1 2 Lim K → + ∞ 2 k + 3 2 k + 1 \u003d 12< 1
A number is convergent.
Example 13.
Determine, is a series of diverging Σ k \u003d 1 ∞ k k k k k k k .
We use the sign of the Dalamber in order to determine the size of the series: Lim K → + ∞ a k + 1 a k \u003d lim k → + ∞ (k + 1) k + 1 (k + 1)! K k k! \u003d Lim K → + ∞ (k + 1) k + 1 · k! K k · (k + 1)! \u003d Lim k → + ∞ (k + 1) k + 1 kk · (k + 1) \u003d \u003d lim k → + ∞ (k + 1) kkk \u003d lim k → + ∞ k + 1 kk \u003d lim k → + ∞ 1 + 1 kk \u003d E\u003e 1
Consequently, the row is divergent.
Cauchy radical sign
Suppose that Σ k \u003d 1 ∞ a k is a sign-positive series. If Lim K → + ∞ A K K< 1 , то ряд является сходящимся, если lim k → + ∞ a k k > 1, then divergent.
Note 2.
If lim k → + ∞ a k k \u003d 1, this feature does not provide any information - additional analysis is required.
This feature can be used in examples that are easy to determine. The case will be characteristic when a member of a numerical series is a significant expression.
In order to secure the information obtained, consider several characteristic examples.
Example 14.
Determine whether the sign is σ k \u003d 1 ∞ 1 (2 k + 1) k on the convergent.
The necessary condition is considered to be made, since Lim K → + ∞ 1 (2 k + 1) k \u003d 1 + ∞ + ∞ \u003d 0.
According to the sign considered above, we obtain Lim K → + ∞ A k k \u003d lim k → + ∞ 1 (2 k + 1) k k \u003d lim k → + ∞ 1 2 k + 1 \u003d 0< 1 . Данный ряд является сходимым.
Example 15.
Are the numeric series σ k \u003d 1 ∞ 1 3 k · 1 + 1 k k 2 converge.
We use the feature described in the previous paragraph Lim K → + ∞ 1 3 k · 1 + 1 k 2 k \u003d 1 3 · Lim k → + ∞ 1 + 1 k \u003d e 3< 1 , следовательно, числовой ряд сходится.
Integral sign Cauchy
Suppose that σ k \u003d 1 ∞ a k is a sign standby. It is necessary to designate the function of continuous argument y \u003d f (x)which coincides with n \u003d f (n). If a y \u003d f (x)more zero, not interrupted and decreases on [A; + ∞), where a ≥ 1
In the event that the immobiliated integral ∫ a + ∞ f (x) D x is converging, then the series in question also converges. If he diverges, in the example of the example, the row also diverges.
When checking the decrease in the function, the material considered on previous lessons can be used.
Example 16.
Consider the example σ k \u003d 2 ∞ 1 k · Ln k for convergence.
The condition of the convergence of the row is considered to be made, since Lim K → + ∞ 1 k · Ln k \u003d 1 + ∞ \u003d 0. Consider y \u003d 1 x · ln x. It is greater than zero, it is not interrupted and decreases at [2; + ∞). The first two points are known to be known, but the third should be stopped in more detail. Find a derivative: y "\u003d 1 x · ln x" \u003d x · ln x "x · ln x 2 \u003d ln x + x · 1 xx · ln x 2 \u003d - ln x + 1 x · ln x 2. It is less than zero On [2; + ∞). This proves the thesis that the function is descending.
Actually, the function y \u003d 1 x · Ln x corresponds to the signs of the principle that we considered above. We use it: ∫ 2 + ∞ dxx · Ln x \u003d Lim A → + ∞ ∫ 2 A d (ln x) Ln x \u003d Lim A → + ∞ ln (ln x) 2 a \u003d \u003d Lim A → + ∞ (ln ( ln a) - ln (ln 2)) \u003d ln (ln (+ ∞)) - ln (ln 2) \u003d + ∞
According to the results obtained, the initial example is divided, since the immutable integral is divergent.
Example 17.
Prove the convergence of the row σ k \u003d 1 ∞ 1 (10 k - 9) (Ln (5 K + 8)) 3.
Since Lim k → + ∞ 1 (10 k - 9) (Ln (5 k + 8)) 3 \u003d 1 + ∞ \u003d 0, then the condition is considered to be made.
Starting with k \u003d 4, the correct expression 1 (10 k - 9) (Ln (5 k + 8)) 3< 1 (5 k + 8) (ln (5 k + 8)) 3 .
If the series Σ k \u003d 4 ∞ 1 (5 k + 8) (Ln (5 K + 8)) 3 will be considered convergent, then, according to one of the principles of comparison, a number σ k \u003d 4 ∞ 1 (10 k - 9) ( LN (5 K + 8)) 3 will also be considered convergent. Thus, we will be able to determine that the initial expression is also converging.
We turn to the proof σ k \u003d 4 ∞ 1 (5 k + 8) (Ln (5 K + 8)) 3.
Since the function y \u003d 1 5 x + 8 (Ln (5 x + 8)) 3 is greater than zero, it is not interrupted and decreases to [4; + ∞). We use the feature described in the previous paragraph:
∫ 4 + ∞ dx (5 x + 8) (Ln (5 x + 8)) 3 \u003d Lim A → + ∞ ∫ 4 A DX (5 x + 8) (Ln (5 x + 8)) 3 \u003d 1 5 · Lim A → + ∞ ∫ 4 A D (Ln (5 x + 8) (Ln (5 x + 8)) 3 \u003d - 1 10 · Lim A → + ∞ 1 (Ln (5 x + 8)) 2 | 4 a \u003d \u003d - 1 10 · Lim A → + ∞ 1 (Ln (5 · a + 8)) 2 - 1 (ln (5 · 4 + 8)) 2 \u003d - 1 10 · 1 + ∞ - 1 (LN 28) 2 \u003d 1 10 · LN 28 2
In the resulting converging series, ∫ 4 + ∞ dx (5 x + 8) (Ln (5 x + 8)) 3, it can be determined that σ k \u003d 4 ∞ 1 (5 k + 8) (Ln (5 k + 8 )) 3 also converges.
Sign of Raabe
Suppose that σ k \u003d 1 ∞ a k is a sign-positive numerical series.
If Lim K → + ∞ k · a k a k + 1< 1 , то ряд расходится, если lim k → + ∞ k · a k a k + 1 - 1 > 1, it converges.
This method of determining can be used if the technique described above does not give visible results.
Research on absolute convergence
For the study we take σ k \u003d 1 ∞ b k. Use the alignmental σ k \u003d 1 ∞ b k. We can use any of the appropriate signs that we described above. If the series Σ k \u003d 1 ∞ b k converges, the initial range is absolutely convergent.
Example 18.
Explore the series Σ k \u003d 1 ∞ (- 1) k 3 k 3 + 2 k - 1 on the convergence of σ k \u003d 1 ∞ (- 1) k 3 k 3 + 2 k - 1 \u003d σ k \u003d 1 ∞ 1 3 k 3 + 2 K - 1.
The condition is performed by Lim K → + ∞ 1 3 k 3 + 2 k - 1 \u003d 1 + ∞ \u003d 0. Use Σ k \u003d 1 ∞ 1 K 3 2 and we use the second feature: Lim K → + ∞ 1 3 k 3 + 2 k - 1 1 K 3 2 \u003d 1 3.
A series Σ k \u003d 1 ∞ (- 1) k 3 k 3 + 2 k - 1 converges. The source row is also absolutely moving.
Divergence of alternate ranks
If a series Σ k \u003d 1 ∞ b k is consigned, then the corresponding alternate range Σ k \u003d 1 ∞ b K is either consigned or conditionally moving.
Only a sign of the Dalamber and the radical sign of Cauchy will help draw the conclusions about σ k \u003d 1 ∞ b k along the divergence of modules σ k \u003d 1 ∞ b k. The series Σ k \u003d 1 ∞ b k is also diverged if the necessary condition of convergence is not fulfilled, that is, if Lim k → ∞ + b k ≠ 0.
Example 19.
Check divergence 1 7, 2 7 2, - 6 7 3, 24 7 4, 120 7 5 - 720 7 6 ,. . . .
Module k-mo A member is represented as b k \u003d k! 7 k.
We explore the series Σ k \u003d 1 ∞ b k \u003d σ k \u003d 1 ∞ k! 7 K for convergence on the basis of the Dalamber: Lim K → + ∞ b K + 1 B k \u003d Lim k → + ∞ (K + 1)! 7 K + 1 K! 7 k \u003d 1 7 · Lim k → + ∞ (k + 1) \u003d + ∞.
Σ k \u003d 1 ∞ b k \u003d Σ k \u003d 1 ∞ k! 7 K diverges the same way as the original option.
Example 20.
Is Σ k \u003d 1 ∞ (- 1) k · k 2 + 1 ln (k + 1) convergent.
Consider on the required condition Lim K → + ∞ bk \u003d lim k → + ∞ k 2 + 1 ln (k + 1) \u003d ∞ ∞ \u003d lim k → + ∞ \u003d k 2 + 1 "(ln (k + 1))" \u003d \u003d Lim k → + ∞ 2 k 1 k + 1 \u003d lim k → + ∞ 2 k (k + 1) \u003d + ∞. The condition is not fulfilled, therefore σ k \u003d 1 ∞ (- 1) k · k 2 + 1 ln (k + 1) row is consigned. The limit was calculated by the rule of Lopital.
Signs for conditional convergence
Sign Leibnitsa
Definition 12.If the members of the alkaline row decrease b 1\u003e b 2\u003e B 3\u003e. . . \u003e. . . And the limit of the module \u003d 0 for k → + ∞, then the range Σ k \u003d 1 ∞ b k converges.
Example 17.
Consider σ k \u003d 1 ∞ (- 1) k 2 k + 1 5 k (k + 1) for convergence.
The series is represented as Σ k \u003d 1 ∞ (- 1) k 2 k + 1 5 k (k + 1) \u003d σ k \u003d 1 ∞ 2 k + 1 5 k (k + 1). The required condition is performed by Lim K → + ∞ \u003d 2 k + 1 5 k (k + 1) \u003d 0. Consider Σ k \u003d 1 ∞ 1 k on the second sign of comparison Lim k → + ∞ 2 k + 1 5 k (k + 1) 1 k \u003d lim k → + ∞ 2 k + 1 5 (k + 1) \u003d 2 5
We obtain that σ k \u003d 1 ∞ (- 1) k 2 k + 1 5 k (k + 1) \u003d σ k \u003d 1 ∞ 2 k + 1 5 k (k + 1) diverges. A series Σ k \u003d 1 ∞ (- 1) k 2 k + 1 5 k (k + 1) converges on the basis of the leibitus: sequence 2 · 1 + 1 5 · 1 · 1 1 + 1 \u003d 3 10, 2 · 2 + 1 5 · 2 · (2 \u200b\u200b+ 1) \u003d 5 30, 2 · 3 + 1 5 · 3 · 3 + 1 ,. . . Lim k → + ∞ \u003d 2 k + 1 5 k (k + 1) \u003d 0.
The row conventionally converges.
Sign of Abel-Dirichlet
Definition 13.Σ k \u003d 1 + ∞ u k · v k converges if (u k) does not increase, and the sequence σ k \u003d 1 + ∞ v k is limited.
Example 17.
Explore 1 - 3 2 + 2 3 + 1 4 - 3 5 + 1 3 + 1 7 - 3 8 + 2 9 +. . . For convergence.
Imagine
1 - 3 2 + 2 3 + 1 4 - 3 5 + 1 3 + 1 7 - 3 8 + 2 9 +. . . \u003d 1 · 1 + 1 2 · (- 3) + 1 3 · 2 + 1 4 · 1 + 1 5 · (- 3) + 1 6 · \u003d σ k \u003d 1 ∞ u k · v k
where (u k) \u003d 1, 1 2, 1 3 ,. . . - inoperative, and the sequence (V k) \u003d 1, - 3, 2, 1, - 3, 2 ,. . . limited (s k) \u003d 1, - 2, 0, 1, - 2, 0 ,. . . . A number converges.
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