Chapter 11 - Infinite Sequences and Series - 11.10 Taylor and Maclaurin Series - 11.10 Exercises - Page 812: 80

$tan^{-1}(\frac{1}{2})$

Given: $\frac{1}{1.2}-\frac{1}{3.2^{3}}+\frac{1}{5.2^{5}}+....=\Sigma_{n=0}^{\infty}(-1)^{n}\dfrac{1}{(2n+1).2^{2n+1}}$ $=\Sigma_{n=0}^{\infty}(-1)^{n}\dfrac{\frac{1}{2^{2n+1}}}{(2n+1)}$ Since, $tan^{-1}x=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{(2n+1)}$ Thus, $tan^{-1}(\frac{1}{2})=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{(1/2)^{2n+1}}{(2n+1)}$