Finding $\tan t$ if $t=\sum_{i=1}^{\infty}\tan^{-1}\bigl(\frac{1}{2i^2}\bigr)$ I am solving this problem.

Problem. If $$\sum_{i=1}^{\infty} \tan^{-1}\biggl(\frac{1}{2i^{2}}\biggr)= t$$ then find the value of $\tan{t}$.

My solution is like the following: I can rewrite:
\begin{align*}
\tan^{-1}\biggl(\frac{1}{2i^{2}}\biggr) & = \tan^{-1}\biggl[\frac{(2i+1) - (2i-1)}{1+(2i+1)\cdot (2i-1)}\biggr] \\\ &= \tan^{-1}(2i+1) - \tan^{-1}(2i-1)
\end{align*}
and when I take the summation the only term which remains is $-\tan^{-1}(1)$, from which I get the value of $\tan{t}$ as $-1$. But the answer appears to be $1$. Can anyone help me on this.
 A: How about trying the same identity, but using the fact that $\tan\left(\frac{\pi}{2}-x\right)=\frac{1}{\tan(x)}$.
$$
\begin{align}
\sum_{k=1}^\infty\tan^{-1}\left(\frac{1}{2k^2}\right)
&=\sum_{k=1}^\infty\tan^{-1}\left(\frac{1}{2k-1}\right)-\tan^{-1}\left(\frac{1}{2k+1}\right)\\
&=\tan^{-1}(1)\\
&=\frac{\pi}{4}
\end{align}
$$
Of course, then $\tan\left(\frac{\pi}{4}\right)=1$
A: In this answer, it is shown that using the equation
$$
\frac1{2k^2}=\frac{\frac1{2k-1}-\frac1{2k+1}}{1+\frac1{2k-1}\frac1{2k+1}}
$$
and the identity
$$
\tan(x-y)=\frac{\tan(x)-\tan(y)}{1+\tan(x)\tan(y)}
$$
we get
$$
\begin{align}
\sum_{k=1}^\infty\tan^{-1}\left(\frac{1}{2k^2}\right)
&=\sum_{k=1}^\infty\tan^{-1}\left(\frac{1}{2k-1}\right)-\tan^{-1}\left(\frac{1}{2k+1}\right)\\
&=\tan^{-1}(1)\\
&=\frac{\pi}{4}
\end{align}
$$
A: HINT:
$$\frac1{2n^2}=\frac2{1+4n^2-1}=\frac{2n+1-(2n-1)}{1+(2n+1)(2n-1)}$$
$$\arctan x-\arctan y=\arctan\left(\frac{x-y}{1+xy}\right)$$
A: By telescopy we deduce
$$\begin{array}{c l}\sum_{k=1}^\infty \tan^{-1}\left(\frac{1}{2k^2}\right) & = \lim_{n\to\infty}\sum_{k=1}^n\tan^{-1}\left(\frac{1}{2k^2}\right) \\ 
& = \lim_{n\to\infty}\sum_{k=1}^n\left[\tan^{-1}(2k+1)-\tan^{-1}(2k-1)\right] \\ 
& =\lim_{n\to\infty}\left[\color{Purple}{\tan^{-1}(2n+1)}-\tan^{-1}(1)\right]. \end{array}$$
However, the term in purple does not tend to zero as $n\to\infty$, it tends to something else...
