<s>[INST] Find the value of $\sum_{m=1}^\infty \tan ^ {-1}\frac{2m}{m^4+m^2+2}$ How to find value of this sum?
$$\sum\limits_{m=1}^\infty \tan^{-1}\left(\frac{2m}{m^4+m^2+2}\right)$$ I can't understand how to simplify this. Should I use any trigonometric substitution to simplify the fraction? Hints and help needed!
 [/INST] \begin{eqnarray}
\frac{2m}{m^4+m^2+2}&=&\frac{2m}{m^4+2m^2-m^2+1+1}\\
&=&\frac{2m}{(m^2+1)^2-m^2+1}\\
&=&\frac{2m}{(m^2+m+1)(m^2-m+1)+1}\\
&=&\frac{(m^2+m+1)-(m^2-m+1)}{(m^2+m+1)(m^2-m+1)+1}
\end{eqnarray}
so almost done! 
$$
\arctan(a)-\arctan(b)=\arctan\left(\frac{a-b}{1+ab}\right)
$$ 
\begin{eqnarray}
\arctan(m^2+m+1)-\arctan(m^2-m+1)&=&\arctan\left(\frac{(m^2+m+1)-(m^2-m+1)}{(m^2+m+1)(m^2-m+1)+1}\right)\\
&=&\arctan\frac{2m}{(m^2+m+1)(m^2-m+1)+1}
\end{eqnarray} 
now you have telescopic sumation, because 
\begin{eqnarray}
\arctan(m^2+m+1)-\arctan(m^2-m+1)&=&\arctan(m(m+1)+1)-\arctan(m(m-1)+1)\\
&=&f(m)-f(m-1)
\end{eqnarray}
$$
\sum_{m=1}^\infty\tan^{-1}\left(\frac{2m}{m^4+m^2+2}\right)=\\\sum_{m=1}^\infty(f(m)-f(m-1))=\\\lim_{m \to \infty}\tan^{-1}(m^2+m+1)-\tan^{-1}1=\\\frac\pi2-\frac\pi4=\\\frac\pi4.
$$
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