Answered: (-1)"(x - 1)2n+I Σ 2n + 1 n=0

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

(a) find the series’ radius and interval of convergence.
Then identify the values of x for which the series converges
(b) absolutely and (c) conditionally.

Expert Solution

Step 1

Given the series $\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} {(x - 1)}^{2 n + 1}}{2 n + 1}$ .

To find the radius and interval of convergence.

(a) Let us first find the radius of convergence of the series.

Here $a_{n} = \frac{{(- 1)}^{n} {(x - 1)}^{2 n + 1}}{2 n + 1}$ and $a_{n + 1} = \frac{{(- 1)}^{n + 1} {(x - 1)}^{2 n + 3}}{2 n + 3}$

Step 2

We know the ratio test for convergence, for the series $\sum_{n = 0}^{\infty} a_{n}$ series converges absolutely if $\lim_{n \to \infty} |\frac{a_{n + 1}}{a_{n}}| < 1$ otherwise diverges.

$\Rightarrow \lim_{n \to \infty} |\frac{a_{n + 1}}{a_{n}}| = \lim_{n \to \infty} |\frac{{(- 1)}^{n + 1} {(x - 1)}^{2 n} \times {(x - 1)}^{3}}{2 n + 3} \times \frac{2 n + 1}{{(- 1)}^{n} {(x - 1)}^{2 n} (x - 1)}|$

$\Rightarrow \lim_{n \to \infty} |\frac{a_{n + 1}}{a_{n}}| = |{(x - 1)}^{2}| \lim_{n \to \infty} |\frac{2 n + 1}{2 n + 3}|$

$\Rightarrow \lim_{n \to \infty} |\frac{a_{n + 1}}{a_{n}}| = |{(x - 1)}^{2}| (1)$

$\Rightarrow \lim_{n \to \infty} |\frac{a_{n + 1}}{a_{n}}| = {(x - 1)}^{2}$

For convergence ${(x - 1)}^{2} < 1$ .

Therefore radius of convergence is $1$ .

Now to find interval of convergence let us solve ${(x - 1)}^{2} < 1 .$

$\Rightarrow - 1 < x - 1 < 1$

$\Rightarrow 0 < x < 2$

Now let us check for endpoints.

For $x = 0$ the series will becomes $\sum_{n = 0}^{\infty} \frac{{(- 1)}^{3 n + 1}}{2 n + 1}$ .

This is an alternating series with $a_{n} = \frac{1}{2 n + 1}$ .

Here $a_{n + 1} < a_{n}$ .

We know the alternating series test the series $\sum_{n = 0}^{\infty} a_{n}$ converges if $a_{n} > 0, a_{n}$ is decreasing and $\lim_{n \to \infty} a_{n} = 0 .$

Now $\lim_{n \to \infty} a_{n} = \lim_{n \to \infty} \frac{1}{2 n + 1} = 0$ .

By alternating series test the series $\sum_{n = 0}^{\infty} \frac{{(- 1)}^{3 n + 1}}{2 n + 1}$ converges.

Now for $x = 2$ the series will becomes $\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1}$ .

This is an alternating series with $a_{n} = \frac{1}{2 n + 1}$ .

Here $a_{n + 1} < a_{n}$ .

Now $\lim_{n \to \infty} a_{n} = \lim_{n \to \infty} \frac{1}{2 n + 1} = 0$ .

By alternating series test the series $\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1}$ converges.

Therefore the interval of convergence of the series $0 \leq x \leq 2 .$

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