Chapter 10.2, Problem 36E

To determine

To find: The converges for the given value of x.

Answer to Problem 36E

At $x=1,$ it converges to ${\tan }^{-1}1=\frac{\pi }{4}$ and at $x=-1,$ it converges to

  ${\tan }^{-1}(-1)=-\frac{\pi }{4}$

Explanation of Solution

Given information:

  $x-\frac{x^3}{3}+\frac{x^5}{5}-\cdots +{(-1)}^n\frac{x^{2n+1}}{2n+1}+\cdots$

Calculation:

The given power series is, $f(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\dots \dots .+{(-1)}^n\frac{x^{2n+1}}{2n+1}+\dots \mathrm{...}$ this series represents

  ${\tan }^{-1}x\cdot$ Therefore, when $x=1,$ it converges to ${\tan }^{-1}1=\frac{\pi }{4}$ when $x=-1,$ it converges to

  ${\tan }^{-1}(-1)=-\frac{\pi }{4}$ .

Therefore, ${\tan }^{-1}1=\frac{\pi }{4}$ at $x=1$ and ${\tan }^{-1}(-1)=-\frac{\pi }{4}$ at $x=-1$ .