# Find $\sum_{n=1}^{\infty}\tan^{-1}\frac{2}{n^2}$

Find $$M:=\sum_{n=1}^{\infty}\tan^{-1}\frac{2}{n^2}$$

There's a solution here that uses complex numbers which I didn't understand and I was wondering if the following is also a correct method.

My proposed solution

\begin{align} &\sum_{n=1}^{\infty}\tan^{-1}\frac{2}{n^2}\\ =&\sum_{n=1}^{\infty}\tan^{-1}\frac{(1+n)+(1-n)}{1-(1+n)(1-n)}\\ =&\sum_{n=1}^{\infty}(\tan^{-1}(1+n)+\tan^{-1}(1-n))\\ =&\sum_{n=1}^{\infty}(\tan^{-1}(n+1)-\tan^{-1}(n-1)) \end{align}

And this implies $$M=\lim_{m\to\infty}(\tan^{-1}(m+1)+\tan^{-1}m-\tan^{-1}1-\tan^{-1}0)=\frac{3\pi}{4}$$

• This may or may not be of help: $\arctan(1/x) = \pi/2 - \arctan(x)$ – user150203 Dec 27 '18 at 5:23
• The 3rd and 4th lines of the equation after "My proposed solution" are missing a summation sign. Otherwise, the solution is correct. – JimmyK4542 Dec 27 '18 at 5:35
• @JimmyK4542 Edited, ty. – Rhaldryn Dec 27 '18 at 5:40
• – lab bhattacharjee Dec 27 '18 at 6:34

## 2 Answers

\begin{align} &\sum_{n=1}^{\infty}\tan^{-1}\frac{2}{n^2}\\ =&\sum_{n=1}^{\infty}\tan^{-1}\frac{(1+n)+(1-n)}{1-(1+n)(1-n)}\\ =\color{red}{\sum_{n=1}^\infty}&\tan^{-1}(1+n)+\tan^{-1}(1-n)\\ =\color{red}{\sum_{n=1}^\infty}&\tan^{-1}(n+1)-\tan^{-1}(n-1) \end{align}

Edit:$$\color{red}{\sum_{n=1}^\infty}$$ was missing in your question before edit. I am not going to delete this.

However your proof is now correct. $$M=\lim_{m\to\infty}(\tan^{-1}(m+1)+\tan^{-1}m-\tan^{-1}1-\tan^{-1}0)=\color{red}{\frac\pi2+\frac\pi2-\frac\pi4-0=\pi-\frac\pi4}=\frac{3\pi}{4}$$

Looks good to me. If I was going to offer a critique I would just say: when writing an argument it's always better to over communicate rather than under communicate.

The first equality is just algebra.

Your second equality requires a little bit to see clearly but it's true. Most will recall:

$$\tan(A+B)=\frac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)}$$ Or if you'd like: $$A+B= \tan^{-1} \bigg(\frac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)} \bigg)$$ Taking $$A=\tan^{-1}(1+n)$$ and $$B=\tan^{-1}(1-n)$$

Honestly adding this much explanation seems like almost overkill.

The 4th equality follows as result of $$\tan^{-1}$$ being an odd function.

Now the last part you are using a telescoping series technique so that you may ignore all the middle terms. That is,

\begin{align} &\sum_{n=1}^\infty\tan^{-1}(n+1)-\tan^{-1}(n-1) \\ &= \lim_{m\to \infty} \tan^{-1}(m+1)-tan^{-1}(m-1)+\dots +\tan^{-1}(4)-\tan^{-1}(2)+\tan^{-1}(3)-\tan^{-1}(1)+tan^{-1}(2)-\tan^{-1}(0) \end{align}

So after we consider what cancels and what doesn't we find that we only need to concern ourselves with $$\lim_{m\to \infty}\tan^{-1}(m+1)+\tan^{-1}(m-1)-\tan^{-1}(1)$$

So while that is true: I think it might merit a sentence or two just to make sure the audience is following.

• Isn't $M=\lim_{m\to \infty} \tan^{-1}(m)-\tan^{-1}(0)=\frac\pi2-0?$ – tatan Dec 27 '18 at 5:50
• @tatan. Yes. I am missing some terms. Let me edit. – Mason Dec 27 '18 at 5:51
• Thanks. I'm really slow at writing in latex so I skip steps. But I will add more explanation in English in the future. – Rhaldryn Dec 27 '18 at 5:57
• Yep. Can I ask you one thing? Are you from Maryland?(I saw from your profile) – tatan Dec 27 '18 at 19:57
• Yes. Uh oh. Stranger danger. :). No. I am not worried. I grew up in Ohio and move to the DC area post college. I went to Ohio University not Univ of Maryland. Though I did cite Dr. Yorke in my undergrad thesis and then I met him wandering once while wandering Univ of Maryland's halls... – Mason Dec 27 '18 at 19:57