Question

If ${\sum }_{n=1}^{\infty }{\tan }^{-1}\left(\frac{18}{n^2}\right)=\frac{k\pi }{\lambda },$ where $k$ and $\lambda$ are natural numbers which are coprime, then value of $|k-\lambda |$ is

Answer 1

$T_n={\tan }^{-1}\left[\frac{2}{1+(\frac{n}{3}+1)(\frac{n}{3}-1)}\right]={\tan }^{-1}(\frac{n}{3}+1)-{\tan }^{-1}(\frac{n}{3}-1)T_1={\tan }^{-1}\left(\frac{4}{3}\right)-{\tan }^{-1}(-\frac{2}{3})T_2={\tan }^{-1}\left(\frac{5}{3}\right)-{\tan }^{-1}(-\frac{1}{3})$
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$T_n={\tan }^{-1}(\frac{n}{3}+1)-{\tan }^{-1}(\frac{n}{3}-1)S=\sum _{n=1}^{\infty }{T}_n=6\times \frac{\pi }{2}-\frac{\pi }{4}S=\frac{11\pi }{4}k=11,\lambda =4$