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

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

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