TITLE: A serie about $\sum_{n = 1}^\infty {\arctan \frac{{10n}}{{\left( {3{n^2} + 2} \right)\left( {9{n^2} - 1} \right)}}}$
QUESTION [6 upvotes]: How to prove $$\sum\limits_{n = 1}^\infty  {\arctan \frac{{10n}}{{\left( {3{n^2} + 2} \right)\left( {9{n^2} - 1} \right)}}}  = \ln 3 - \frac{\pi }{4}.$$
Add: Maybe we can follow this!

REPLY [9 votes]: Since $\arctan(x)=\arg(1+ix)$ and we can factor
$$
1+\frac{10in}{\left(3n^2+2\right)\left( 9n^2-1\right)}
=\frac{\left(1-\frac in\right)\left(1+\frac i{3n-1}\right)\left(1+\frac i{3n+1}\right)\left(1+\frac i{3n}\right)}{1+\frac2{3n^2}}
$$
we have that
$$
\begin{align}
&\arctan\left(\frac{10n}{\left(3n^2+2\right)\left( 9n^2-1\right)}\right)\\[6pt]
&=\arctan\left(\frac1{3n-1}\right)+\arctan\left(\frac1{3n}\right)+\arctan\left(\frac1{3n+1}\right)-\arctan\left(\frac1n\right)
\end{align}
$$
Now this becomes a telescoping series:
$$
\begin{align}
&\sum_{n=1}^\infty\arctan\left(\frac{10n}{\left(3n^2+2\right)\left( 9n^2-1\right)}\right)\\
&=\lim_{m\to\infty}\sum_{n=1}^m\left[\arctan\left(\frac1{3n-1}\right)+\arctan\left(\frac1{3n}\right)+\arctan\left(\frac1{3n+1}\right)-\arctan\left(\frac1n\right)\right]\\
&=-\arctan(1)+\lim_{m\to\infty}\sum_{n=m+1}^{3m+1}\arctan\left(\frac1n\right)\\
&=-\arctan(1)+\lim_{m\to\infty}\sum_{n=m+1}^{3m+1}\left[\frac1n+O\left(\frac1{n^3}\right)\right]\\[6pt]
&=\log(3)-\frac\pi4
\end{align}
$$