# 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.

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. May 14, 2012 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, 2012 at 14:02
• FWIW, I can barely see the distinction between black and purple on my screen, and it is generally unwise to indicate information with color alone because enough of the human population is colorblind. (Not me, I'm just old.) It's okay to duplicate information with color (making it clear with text what you are talking about, but making it visually striking to the readers who can see the color) but it should rarely be the only way you present information. May 9, 2017 at 16:21

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

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 Apr 20, 2014 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) Apr 20, 2014 at 19:30

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 Apr 20, 2014 at 15:44
• @user34304, Observation probably the best word to respond Apr 20, 2014 at 15:45
• Okay, a Better way to put it, why did u make that manipulation? Apr 20, 2014 at 15:46
• @user34304, to form an expression of the form $$\frac{x-y}{1+xy}$$ Apr 20, 2014 at 15:46
• What is telescopy, sir? Apr 20, 2014 at 15:50