Question

Sum the following series:
${\tan }^{-1}\frac{1}{3}+{\tan }^{-1}\frac{2}{9}+{\tan }^{-1}\frac{4}{33}+...+{\tan }^{-1}\frac{2^{n-1}}{1+2^{2n-1}}$

Answer 1

$$\begin{gathered} {\tan }^{-1}\frac{1}{3}+{\tan }^{-1}\frac{2}{9}+{\tan }^{-1}\frac{4}{33}+...+{\tan }^{-1}\frac{2^{n-1}}{1+2^{2n-1}} \\ \Rightarrow {\tan }^{-1}\left(\frac{2-1}{1+2\times 1}\right)+{\tan }^{-1}\left(\frac{4-2}{1+4\times 2}\right)+{\tan }^{-1}\left(\frac{8-4}{1+8\times 4}\right)+...+{\tan }^{-1}\left(\frac{2^n-2^{n-1}}{1+2^n.2^{n-1}}\right) \\ \Rightarrow \left({\tan }^{-1}2-{\tan }^{-1}1\right)+\left({\tan }^{-1}4-{\tan }^{-1}2\right)+\left({\tan }^{-1}8-{\tan }^{-1}4\right)+...+\left({\tan }^{-1}{2}^{n-1}-{\tan }^{-1}{2}^{n-2}\right)+\left({\tan }^{-1}{2}^n-{\tan }^{-1}{2}^{n-1}\right) \\ \Rightarrow {\tan }^{-1}{2}^n-{\tan }^{-1}1 \\ \Rightarrow {\tan }^{-1}{2}^n-\frac{\mathrm{\pi }}{4} \end{gathered}$$