Answered: (n + 1)(2x + 1)" (2n + 1)2" 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 $\sum_{n = 0}^{\infty} \frac{(n + 1) {(2 x + 1)}^{n}}{(2 n + 1) 2^{n}}$ .

To find the radius and interval of convergence and also to find the values of $x$ for which the series converges absolutely and converges conditionally.

The radius of convergence $R$ is any number such that the will converge for $|x - a| < R$ and diverge for $|x - a| > R$ .

Step 2

(a) Let us first find radius and interval of convergence of given series.

Here $\sum_{n = 0}^{\infty} \frac{(n + 1) {(2 x + 1)}^{n}}{(2 n + 1) 2^{n}}$ .

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

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

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

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

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

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

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

Since by using ratio test $|\frac{(2 x + 1)}{2}| < 1 .$

Therefor radius of convergence $R = 1$ .

$\Rightarrow |2 x + 1| < 2$

$\Rightarrow - 2 < 2 x + 1 < 2$

$\Rightarrow - 2 - 1 < 2 x < 2 - 1$

$\Rightarrow - 3 < 2 x < 1$

$\Rightarrow - \frac{3}{2} < x < \frac{1}{2}$

Let us check for endpoints.

At $x = - \frac{3}{2}$ given series will be $\sum_{n = 0}^{\infty} \frac{(n + 1) {(- 2)}^{n}}{(2 n + 1) 2^{n}} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} (n + 1) {(2)}^{n}}{(2 n + 1) 2^{n}} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} (n + 1)}{(2 n + 1)}$

We know that the series divergence test if the $\lim_{n \to \infty} a_{n} \neq 0$ then the series diverges.

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

Thus the series diverges at $x = - \frac{3}{2}$ .

Now at $x = \frac{1}{2}$ given series will be $\sum_{n = 0}^{\infty} \frac{(n + 1) {(2)}^{n}}{(2 n + 1) 2^{n}} = \sum_{n = 0}^{\infty} \frac{(n + 1)}{(2 n + 1)}$ .

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

Clearly by series divergence test the series diverges at $x = \frac{1}{2}$ .

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