Question

Find $\frac{8}{\pi }{\sum }_{k=1}^{\infty }{\tan }^{-1}\left(\frac{2k}{2+k^2+k^4}\right)$

Answer 1

${\sum }_{k=1}^n{\tan }^{-1}\frac{2k}{1+(k^2+k+1)(k^2-k+1)}$
${\sum }_{k=1}^n{\tan }^{-1}\frac{(k^2+k+1)-(k^2-k+1)}{1+(k^2+k+1)(k^2-k+1)}$
$\Rightarrow {\sum }_{k=1}^n\left({\tan }^{-1}\right(k^2+k+1)-{\tan }^{-1}(k^2-k+1\left)\right)$
$={\tan }^{-1}(n^2+n+1)-{\tan }^{-1}1$
When $n\to \infty$
Then summation ${\sum }_{k=1}^{\infty }ta{n}^{-1}\frac{2k}{2+k^2+k^4}=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$
Hence $\frac{8}{\pi }\times \frac{\pi }{4}$ is $2$