Evaluate Infinite Sum for $\tan a$

The problem asks for the value of $\tan a$, where $a$ is defined by the infinite sum: $a = \sum_{r=1}^{\infty} \tan^{-1}\left(\frac{1}{2r^2}\right)$

Transform the Series Term

We can rewrite the general term $\tan^{-1}\left(\frac{1}{2r^2}\right)$ using the arctan difference identity: $ \tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x-y}{1+xy}\right) $ Let's find suitable expressions for $x$ and $y$. Consider $x = 2r+1$ and $y = 2r-1$. Then: $ x - y = (2r+1) - (2r-1) = 2 $ $ 1 + xy = 1 + (2r+1)(2r-1) = 1 + (4r^2 - 1) = 4r^2 $ So, $ \tan^{-1}\left(\frac{x-y}{1+xy}\right) = \tan^{-1}\left(\frac{2}{4r^2}\right) = \tan^{-1}\left(\frac{1}{2r^2}\right) $ Therefore, the general term can be expressed as: $ \tan^{-1}\left(\frac{1}{2r^2}\right) = \tan^{-1}(2r+1) - \tan^{-1}(2r-1) $

Apply Telescoping Series Method

The infinite sum $a$ becomes a telescoping series: $ a = \sum_{r=1}^{\infty} [\tan^{-1}(2r+1) - \tan^{-1}(2r-1)] $ Let's examine the partial sum $S_N$: $ S_N = \sum_{r=1}^{N} [\tan^{-1}(2r+1) - \tan^{-1}(2r-1)] $ Writing out the first few terms and the Nth term: $ S_N = [\tan^{-1}(3) - \tan^{-1}(1)] \quad (r=1) $ $ + [\tan^{-1}(5) - \tan^{-1}(3)] \quad (r=2) $ $ + [\tan^{-1}(7) - \tan^{-1}(5)] \quad (r=3) $ $ + \dots $ $ + [\tan^{-1}(2N+1) - \tan^{-1}(2N-1)] \quad (r=N) $ The intermediate terms cancel out, leaving: $ S_N = \tan^{-1}(2N+1) - \tan^{-1}(1) $

Calculate the Sum Value $a$

To find the value of $a$, we take the limit of the partial sum $S_N$ as $N$ approaches infinity: $ a = \lim_{N \to \infty} S_N = \lim_{N \to \infty} [\tan^{-1}(2N+1) - \tan^{-1}(1)] $ We know that $\lim_{x \to \infty} \tan^{-1}(x) = \frac{\pi}{2}$ and $\tan^{-1}(1) = \frac{\pi}{4}$. So, $ a = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} $

Compute $\tan a$

Finally, we need to compute $\tan a$: $ \tan a = \tan\left(\frac{\pi}{4}\right) $ $ \tan a = 1 $