Answered: g(x) = LI+ 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

Define

Find the values of x where the series converges and show that we get a continuous function on this set.

Expert Solution

Step 1

The given series is $g (x) = \sum_{n = 0}^{\infty} \frac{x^{2 n}}{1 + x^{2 n}}$ .

To Find: The values of x for which the series is convergent.

To Prove: The series is convergent to a continuous function on this set.

Step 2

The given series is,

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

According to Ratio test,

If $\lim_{n \to \infty} |\frac{a_{n + 1}}{a_{n}}| = L < 1$ , then the series is absolutely convergent.

$\lim_{n \to \infty} |\frac{x^{2 n + 2}}{1 + x^{2 n + 2}} \times \frac{1 + x^{2 n}}{x^{2 n}}| = \lim_{n \to \infty} |\frac{x^{2}}{1 + x^{2 n + 2}} \times \frac{1 + x^{2 n}}{1}| < 0$

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

Thus, the series is convergent when $|x| < 1$ .

At x=1,

$g (x) = \sum_{n = 0}^{\infty} \frac{x^{2 n}}{1 + x^{2 n}} = \sum_{n = 0}^{\infty} \frac{1}{2}$ which is clearly divergent.

Therefore, the series is convergent only when $|x| < 1$ .

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