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 
$$M=\lim_{m\to\infty}(\tan^{-1}(m+1)+\tan^{-1}m-\tan^{-1}1-\tan^{-1}0)=\frac{3\pi}{4}$$
 A: $$\begin{align}
&\sum_{n=1}^{\infty}\tan^{-1}\frac{2}{n^2}\\
=&\sum_{n=1}^{\infty}\tan^{-1}\frac{(1+n)+(1-n)}{1-(1+n)(1-n)}\\
=\color{red}{\sum_{n=1}^\infty}&\tan^{-1}(1+n)+\tan^{-1}(1-n)\\
=\color{red}{\sum_{n=1}^\infty}&\tan^{-1}(n+1)-\tan^{-1}(n-1)
\end{align}
$$
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.
$$M=\lim_{m\to\infty}(\tan^{-1}(m+1)+\tan^{-1}m-\tan^{-1}1-\tan^{-1}0)=\color{red}{\frac\pi2+\frac\pi2-\frac\pi4-0=\pi-\frac\pi4}=\frac{3\pi}{4}$$
A: 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. 
