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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!

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\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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  • $\begingroup$ thank you for comment and correction ! $\endgroup$ – Khosrotash Apr 14 '15 at 21:13
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    $\begingroup$ Very nice work! +1 $\endgroup$ – Nicolas Apr 14 '15 at 21:15

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