Higher_Mathematics_Part_3
.pdf1.4. Tests for Convergence of Positive Terms Series
Comparison tests for positive terms series make it possible to define whether a number series is convergent (divergent) or not by comparing it with another series that is known as convergent (divergent).
Theorem 7 |
(Comparison test for positive terms series). Let |
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∞ |
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a1 + a2 + ... + an + ... = ∑ an , |
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(1.8) |
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n=1 |
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∞ |
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b1 + b2 + ... + bn + ... = ∑ bn , |
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(1.9) |
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n=1 |
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be two series of positive terms, so that |
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0 ≤ an ≤ bn |
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(1.10) |
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∞ |
∞ |
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for all n. Then, if |
∑ bn converges, |
∑ an converges as well; and if |
∑ an di- |
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n=1 |
n=1 |
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n=1 |
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verges, then ∑ bn |
diverges as well. |
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n=1 |
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Proof. We form the partial sums of (1.8) and (1.9) |
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Sn = a1 + a2 + ... + an , σn = b1 + b2 + ...+ bn . |
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It follows from (1.10) that Sn ≤ σn |
for all n = 1, 2, .... |
lim σn |
= σ of its |
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(1) Suppose that series (1.9) converges, i. e, there exists |
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n→∞ |
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partial sum. Since the terms of these series are positive, then |
0 < σn < σ, and it |
follows by (1.10) that 0 < Sn < σ for n = 1, 2, ... .
Thus, all the partial sums Sn of (1.8) are bounded and increase with n, since an > 0 for all n. Consequently, the sequence of partial sums {Sn } is convergent,
i. e., there exists lim Sn = S, which implies that |
∞ |
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∑ an is a convergent series. |
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n→∞ |
n=1 |
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Now, from the inequality 0 < Sn < σ, which holds for all natural n, we ob- |
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tain the inequality |
0 ≤ S ≤ σ if n → ∞, i. e., the sum S of (1.8) does not exceed |
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the sum σ of the convergent series (1.9). |
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∞ |
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(2) Suppose that ∑an diverges. Since all an > 0, then Sn increases with n, |
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and hence lim Sn |
n=1 |
σn ≥ Sn (n = 1, 2, ...) we get |
= + ∞. From the inequality |
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n→∞ |
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lim σn = + ∞, i. e., |
∞ |
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∑ bn diverges. |
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n→∞ |
n=1 |
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Remark 1. Theorem 7 is valid even when (1.10) holds not for all n, but only
begins with some k, i.e., for all n ≥ k, since when we drop a finite number of terms in the beginning, we do not violate the convergence of the series.
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Remark 2. Theorem 7 is also valid for a more general inequality an ≤ λbn (λan ≤ bn ) (n = 1, 2, ...), where (λ > 0).
Theorem 8 (Comparison test for positive terms series). If there exists a finite nonzero
lim |
an |
= L |
(0 < L < +∞), |
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n→∞ b |
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n |
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then (1.8) and (1.9) converge or diverge simultaneously.
Proof. Indeed, from the existence of the above limit it follows that for any ε > 0 is subjected to the condition L — ε > 0, there exists a number N such that
for all n > N
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an |
− L |
< ε. |
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b |
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an |
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n |
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or L − ε < |
< L + ε. |
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b |
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n |
< (L + ε)bn |
for n > N. |
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Hence (L − ε)bn < an |
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∞ |
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If (1.9) converges, |
so does |
∑ (L + ε)bn . But since an < (L + ε)bn |
for |
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n=1 |
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n > N, then, by Theorem 7, series (1.8) will converge as well. |
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∞ |
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< an |
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If (1.9) diverges then |
∑ (L − ε)bn diverges as well. And since (L − ε)bn |
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n=1 |
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for n > N, then, by Theorem 7, series (1.8) also diverges.
D’Alembert’s test for convergence of a series is given by the following theorem.
∞
Theorem 9 (D’Alembert’s test). Consider ∑ an , where an > 0. If there exists
n=1
lim an+1 = l,
n→∞ an
then for 0 ≤ l < 1 the series converges, and for l > 1 the series diverges.
Proof. Suppose that there exists the finite limit lim |
an+1 |
= l, where |
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n→∞ |
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n |
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l < 1. Take any number q so that l < q < 1. Then, for any ε > 0, e. g., for ε = q – l, there exists a number N such that for all n ≥ N we will have
an + 1 −l < q – l. an
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Specifically, we will have |
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an+1 |
−l < q – l or |
an+1 |
< q. |
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a |
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n |
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n |
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Hence an+1 < an q for all n ≥ N. From the latter inequality we will obtain, if n assumes the values N, N + 1, N + 2, ...
aN +1 < aN q,
aN + 2 < aN +1q < aN q2 , aN + 3 < aN + 2q < aN q3 ,
.................................
So, the terms of the series
aN +1 + aN + 2 + aN + 3 + ...
are not larger than the corresponding terms of the series aN q + aN q2 + aN q3 + ...,
which converges as a series of the terms of a geometric progression with the ratio
q, where 0 < q < 1. According to comparison test, the series |
aN +1 + aN + 2 + |
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∞ |
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+aN + 3 + ... converges, and hence ∑ an converges too. |
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n=1 |
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If l > 1, then beginning with a certain number N we will have |
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aN+1 |
>1 or aN +1 > aN > 0. |
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aN |
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Consequently, lim an ≠ 0 |
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∞ |
diverges, since the required |
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and the series ∑ an |
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n→∞ |
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n=1 |
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test for convergence is not satisfied. |
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Remark. If lim |
an+1 |
= 1, |
then the D’Alembert’s test gives no answer as to |
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n→∞ a |
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n |
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whether a series converges or not. |
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∞ |
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Theorem 10 |
(Cauchy’s test). Consider the series |
∑ an , an > 0, |
n = 1, 2, ... . If |
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there exists a finite |
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n=1 |
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lim n a = λ, |
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n→∞ |
n |
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then for 0 ≤ λ < 1 the series converges, and for λ > 1 diverges. |
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Proof. (1) Let λ < 1 . We take q |
such that λ < q < 1. Since there exists |
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lim n a = λ, where λ < q, |
then beginning with a certain number N we will |
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n→∞ |
n |
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≥ N. And so the terms from aN +1 are |
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have |
n an = q, |
whence an |
= qn for n |
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∞ |
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By the |
smaller than the corresponding terms of the convergent series ∑ qn . |
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n= N |
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∞ |
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∞ |
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comparison test |
∑ qn converges, and hence ∑ an converges too. |
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n= N |
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n=1 |
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(2) Let λ > 1. |
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Then, beginning with a certain N and for all n > N, we will have |
n a |
n |
> 1 or |
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an > 1. Accordingly, lim an |
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∞ |
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≠ 0 and the series ∑ an diverges. |
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n→∞ |
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n=1 |
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∞ |
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Remark. If λ = 1, then |
∑ an can either converge or diverge. |
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n=1 |
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∞ |
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Theorem 11 |
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(Integral test). Let a series ∑ an be positive, i.e. an |
≥ 0. Let a |
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n=1 |
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≥ 1 so |
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function f(x) be defined, continuous, positive and nonincreasing for x |
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that f (n) = an . Then the number series |
∞ |
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∑ an converges, if and only if the inte- |
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n=1 |
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+∞ |
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∞ |
+∞ |
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gral |
∫ f (x)dx converges and ∑an ≤ a1 |
+ ∫ f (x)dx. |
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1 |
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n=1 |
1 |
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Remark. The theorem also holds for x ≥ a, |
where a is any number larger |
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than unity. |
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Example. Examine |
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∑ 1p |
= 1+ 1p |
+ 1p + ... + 1p + ... |
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∞ |
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i=1 n |
2 |
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n |
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for convergence.
Using the Integral test, we shall investigate the behaviour of this series. We
have f (n) = |
1 |
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the former satisfying the condition of Integral test. The im- |
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n p |
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1 |
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∞ |
dx is known to converge for p > 1 and diverge for p ≤ 1. |
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proper integral |
∫ |
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p |
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1 |
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∞ |
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Hence, |
the |
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series ∑ |
1 |
converges when p > 1 and diverges when |
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p |
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p ≤ 1. |
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i=1 n |
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Specifically, |
if p = 1 we |
will have the harmonic series |
1+ |
1 |
+ |
1 |
+ ... |
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3 |
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∞ |
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... + |
+ ... = ∑ |
1 |
, and diverges which has already been shown. |
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n |
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n=1 n |
+∞ |
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Remark. In the integral ∫ |
f (x)dx the lower limit may be taken arbitrary, |
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1 |
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e.g., m, where m ≥ 1. |
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1.5. Alternating Series. Leibniz’ Test |
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Definition. The number series |
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∞ |
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a1 − a2 + a3 − ... + (−1)n−1 an + ... = ∑ (−1)n+1 an |
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(1.11) |
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n=1 |
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where an > 0 (n = 1, 2, ...) is called the alternating series. For example, the series 1− 12 + 13 − 14 + ... .
is an alternating one.
The following test, known as the Leibniz’s test, holds for alternating series.
Theorem 12 |
(Leibniz’ test). Suppose that in the alternating series (1.11) all |
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an are such that |
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1) |
a1 > a2 |
> a3 > ... > an > ... |
and |
lim an |
= 0. |
2) |
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n→∞ |
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Then, the series converges, its sum S is positive and does not exceed the first
term, i.e., 0 < S ≤ a1.
Proof. We take the even partial sum S2n and write it as
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S2n = a1 − a2 + a3 − a4 + ... + a2n−1 − a2n = |
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= (a1 − a2 ) + (a3 − a4 ) + ... + (a2n−1 − a2n ). |
where each difference is positive (from the given condition 1). It follows that |
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S2n ≤ a1 − (a2 − a3 ) − (a4 − a5 ) − ... − (a2n− 2 − a2n−1 ) − a2n ≤ a1. |
The |
sequence {S2n } thus increases monotonically and is bounded i.e., |
0 < S2n |
≤ a1 for all n. |
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Consequently, it has the limit |
lim S2n |
= S so that 0 < S ≤ a1. |
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n→∞ |
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The odd partial sum S2n+1 will be |
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S2n+1 = S2n + a2n+1 |
(n = 1, 2, ...). |
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We have proved that lim S2n |
= S, i. e., the series converges. In particular, |
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n→∞ |
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from the inequality 0 < S ≤ a1 it follows that the sum of the series is positive. Remark. The theorem is valid if the condition at which {S2n } is monotonic
is met starting with a certain N for all n ≥ N, so that the discarding of a finite number of terms does not change the convergence of the series.
1.6. Absolute and Conditional Convergence
Let’s consider the series
a1 + a2 + a3 + ... + an + ...
and prove the following theorem:
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∞ |
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∞ |
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Theorem 13 |
If the series |
∑ |
an |
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converges, then the series ∑ an converges as |
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well. |
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n=1 |
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n=1 |
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Proof. From the inequality − |
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an |
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≤ an ≤ |
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an |
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we get 0 ≤ an + |
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an |
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≤ 2 |
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an |
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for |
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(n = 1, 2, ...). |
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∞ |
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∞ |
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Let the series ∑ |
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an |
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be convergent, then the series |
∑ 2 |
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an |
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will be conver- |
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n=1 |
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∞ |
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gent, and from the comparison test the series |
∑ (an + |
an |
) will be convergent |
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∞ |
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n=1 |
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too. But ∑ an |
is the difference of two convergent series |
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n=1 |
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∞ |
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∞ |
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∞ |
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∑ |
(an + |
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an |
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an |
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= ∑ an , |
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n=1 |
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therefore it will converge. |
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Corollary. If ∑ |
an |
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converges, then we have |
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≤ ∑ |
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n=1 |
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∞ |
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When examining ∑ |
an |
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for convergence we can make use of all sufficient |
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tests established for series of positive terms. |
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16 |
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http://vk.com/studentu_tk, http://studentu.tk/
∞
Remark. Generally speaking, the convergence of ∑ an does not suggest the
n=1
∞
convergence of ∑ an , i.e. the theorem only gives a sufficient condition for
n=1
∞
convergence of ∑ an .
n=1
∞
Definition. The series of positive and negative terms ∑ an is called abso-
n=1
∞
lutely convergent, if the series ∑ an converges.
n=1
∞
The series ∑ an is called conditionally convergent if it converges and the
n=1
∞
series ∑ an diverges.
n=1
Micromodule 1
EXAMPLES OF PROBLEMS SOLUTION
Example 1. Prove that the series
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+ ...+ |
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1 |
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1 |
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∞ |
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1 2 |
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2 3 |
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n(n +1) |
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n=1 |
n(n +1) |
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converges. |
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Solution. We consider the nth partial sum of the series |
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Sn = 1 |
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1 . |
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1 2 |
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2 3 |
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we represent Sn in such form |
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Using the usual relation |
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S = |
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1 2 2 3 3 4 |
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= 1 |
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+1 |
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Passing to the limit as n → ∞, |
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lim Sn |
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= lim 1 |
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= 1. |
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n→∞ |
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n→∞ |
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By the definition the series converges and its sum is S = 1.
17
http://vk.com/studentu_tk, http://studentu.tk/
Example 2. Prove that the series
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+...+ |
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1 |
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+... = ∑ |
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1 4 |
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4 7 |
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7 10 |
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(3n− 2)(3n+1) |
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converges. |
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Solution. We represent Sn |
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4 7 |
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Passing to the limit as n → ∞, |
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By the definition |
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converges |
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its |
sum |
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is |
S = |
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or |
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n=1 |
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Example 3. The number series |
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2 + cos π + ... = ∑ cos π |
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−1+ 0 + |
1 + |
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∞ |
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5 |
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diverges, since |
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= lim cos π = cos0 = 1 ≠ 0 (corollary of theorem 6). |
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n→∞ |
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n→∞ |
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Example 4. The series |
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∞ |
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1− 1+ 1− |
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1+ ... = ∑ (−1)n+1 |
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= lim(−1)n+1 |
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n=1 |
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diverges, since |
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lim a |
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does not exist. |
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n→∞ |
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n→∞ |
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∞ |
1 |
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Example 5. Investigate the number series ∑ |
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for convergence. |
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n |
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n=1 |
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Solution. We have |
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lim a |
= lim |
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= 0 but the given series diverges. |
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n→∞ |
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n→∞ |
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Let’s consider the partial sum |
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Sn = 1+ |
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+ ... + |
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> |
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= n |
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then |
Sn > |
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n . |
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http://vk.com/studentu_tk, http://studentu.tk/
Passing to the limit we obtain
lim Sn = ∞.
n→∞
Hence the given series diverges. Example 6. Examine the series
|
1+ 5 + |
7 + ... + 2n + 1 + ... = ∑ 2n + 1 |
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∞ |
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4 5 |
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n + 2 |
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n=1 |
n + 2 |
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for convergence |
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2n + 1 |
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Solution. We have |
lim a |
= lim |
= 2 ≠ 0. |
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n→∞ |
n |
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n→∞ |
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n + 2 |
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Then the given series is divergent. |
1 |
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∞ |
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Example 7. Examine the series ∑ |
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for convergence. |
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Solution. We have |
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n=1 2n + |
n |
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(n = 0, 1, 2 ...). |
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≤ |
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Since ∑ |
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converges, then by the comparison test the original series |
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n=1 2 |
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converges as well. |
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∞ |
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Example 8. Examine the series ∑ |
1 |
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for convergence. |
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n= 2 ln n |
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1 |
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Solution. From the inequality ln n < n follows |
> |
for n = 2, 3, ..., and |
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ln n |
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∞ |
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∞ |
1 |
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the harmonic series |
∑ |
1 |
diverges (in this case |
∑ |
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diverges too). Then by the |
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n=1 n |
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n= 2 n |
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comparison test the original series diverges. |
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∞ |
π |
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Example 9. Examine the series ∑(1− cos |
) for convergence. |
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n |
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n=1 |
2 |
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Solution. Using the inequality sin x < x, which holds for all x ≥ 0, we find
that
0 < 1− cos |
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π |
= |
2sin |
2 |
π |
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≤ 2( |
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π |
) |
2 |
= |
π2 |
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2n |
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2 2n |
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2n |
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2 |
4n |
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π |
2 |
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∞ |
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for n = 1, 2, ... . Here λ = |
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and |
∑ |
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— converges. By comparison test |
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2 |
n |
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n=1 |
4 |
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(and considering Remark 2), the original series converges. |
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19 |
http://vk.com/studentu_tk, http://studentu.tk/
∞ |
π |
for convergence. |
Example 10. Examine the series ∑ sin |
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n=1 |
n |
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∞ 1
Solution. We compare the given series with the harmonic series ∑ and have
n=1 n
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π |
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lim |
an |
= lim |
sin n |
= π 1 = π ≠ 0. |
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1 |
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n→∞ bn |
n→∞ |
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n
The harmonic series diverges, therefore, by the corollary, the original series diverges too.
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∞ |
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1 |
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Example 11. Examine the series ∑ |
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for convergence. |
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2 |
n |
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− n |
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n=1 |
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Solution. We take for comparison the convergent series |
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Then |
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n=1 |
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an |
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lim |
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2n − n |
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= 1 ≠ |
0. |
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n→∞ bn |
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n→∞ |
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n→∞ |
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2n |
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2n |
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since lim |
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n→∞ 2n |
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Example 12. Examine the series ∑ |
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n2 |
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n=1 |
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Solution. We have a |
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2n |
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(n + 1)2 2n |
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1 |
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lim |
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2n+1 n2 |
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n→∞ |
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n→∞ |
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n→∞ 2 |
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By the D’Alembert’s test, the series converges.
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∞ |
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Example 13. Examine the series ∑ |
n |
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for convergence. |
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nn |
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n=1 |
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(n + 1)n+1 |
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(n + 1)n |
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Solution. We have |
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and a |
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n |
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n! |
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n+1 |
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(n + 1)! |
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n! |
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a |
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(n + 1)n n! |
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1 |
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lim |
n |
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= lim |
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= lim(1+ |
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a |
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n!nn |
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n |
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n→∞ |
n |
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n→∞ |
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n→∞ |
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By the D’Alembert’s test, the series diverges.
20
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