# 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.

-

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...

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

-

HINT:

$$\frac1{2n^2}=\frac2{1+4n^2-1}=\frac{2n+1-(2n-1)}{1+(2n+1)(2n-1)}$$

$$\arctan x-\arctan y=\arctan\left(\frac{x-y}{1+xy}\right)$$

-
How did u think of that manipulation in the first step – user34304 Apr 20 '14 at 15:44
@user34304, Observation probably the best word to respond – lab bhattacharjee Apr 20 '14 at 15:45
Okay, a Better way to put it, why did u make that manipulation? – user34304 Apr 20 '14 at 15:46
@user34304, to form an expression of the form $$\frac{x-y}{1+xy}$$ – lab bhattacharjee Apr 20 '14 at 15:46
What is telescopy, sir? – user34304 Apr 20 '14 at 15:50

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}

-
I didn't get how u split the expression – user34304 Apr 20 '14 at 15:45
Really, you should be stopping at the step $\tan^{-1}1$, since the question was to find $\tan t$, which is $1$. (i.e. the calculation $\tan^{-1}1=\frac\pi4$ is irrelevant) – Mario Carneiro Apr 20 '14 at 19:30