22. Evaluate $\sum_{k=0}^{\infty} \frac{2}{\pi} \tan ^{-1}\left(\frac{2}{(2 k+1)^{2}}\right)$.
22. Answer: 1
$$
\begin{array}{l}
\sum_{k=0}^{\infty} \frac{2}{\pi} \tan ^{-1}\left(\frac{2}{(2 k+1)^{2}}\right) \\
=\lim _{n \rightarrow \infty} \sum_{k=0}^{n} \frac{2}{\pi} \tan ^{-1}\left(\frac{2}{(2 k+1)^{2}}\right) \\
=\lim _{n \rightarrow \infty} \sum_{k=0}^{n} \frac{2}{\pi} \tan ^{-1}\left(\frac{2 k+2-2 k}{1+(2 k)(2 k+2)}\right) \\
=\lim _{n \rightarrow \infty} \sum_{k=0}^{n} \frac{2}{\pi}\left(\tan ^{-1}(2 k+2)-\tan ^{-1}(2 k)\right) \\
=\lim _{n \rightarrow \infty} \frac{2}{\pi} \tan ^{-1}(2 n+2)=1 .
\end{array}
$$