# Finding $\tan t$ if $t=\sum\tan^{-1}(1/2t^2)$

I am solving this problem. If $$\sum\limits_{i=1}^{\infty} \tan^{-1}\biggl(\frac{1}{2i^{2}}\biggr)= t$$ 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.

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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...

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Dear Anon, I don't understand your last comment. Why should the purple term tend to $0$. Moreover, that term $\tan^{-1}(2n+1)$ will get cancelled with the $\tan^{-1}(2(n+1)-1)$th term since $n$ here $\to \infty$. So how will $\tan^{-1}(2n+1)$ term remain. –  pattinson May 14 '12 at 13:43
@pattinson: Yes, it will cancel with $\tan^{-1}(2(n+1)-1)$, but then you will be left with a $\tan^{-1}(2(n+1)+1)$ term! The $n$ simply shifts up one after cancellation, it does not disappear. There will always be a final term in the telescopy. Compare with $$\left(\frac{1}{2}-0\right)+\left(\frac{2}{3}-\frac{1}{2}\right)+\left(\frac{3}‌​{4}-\frac{2}{3}\right)+\cdots=\lim_{n\to\infty}\left(\color{Purple}{\frac{n}{n+1}‌​}-0\right)=1.$$ –  anon May 14 '12 at 14:02

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 $\tan\left(\frac{\pi}{4}\right)=1$

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