<s>[INST] On the summation $\sum \limits_{n=1}^{\infty} \arctan \left ( \frac{1}{n^3+n^2+n+1} \right )$ Here is a problem that I ran into. I seriously doubt if there is a closed form but you never know.
Evaluate the series
$$\mathcal{S} = \sum_{n=1}^\infty \arctan \left ( \frac 1 {n^3+n^2+n+1} \right) $$
I searched in vain to attack it using telescopic summation but I failed miserably. Then I remembered the following technique. Since ${\rm Im} \log (1+ix) = \arctan x$ we can express the sum as follows
\begin{align*}
\sum_{n=1}^\infty \arctan \left ( \frac{1}{n^3+n^2+n+1} \right ) &= \sum_{n=1}^\infty \arctan \left [ \frac{1}{(n+1)(n^2+1)} \right ] \\ 
 &= \sum_{n=1}^\infty \operatorname{Im} \left [ \log \left ( 1 + \frac{i}{(n+1)(n^2+1)} \right ) \right ] \\ 
 &= \operatorname{Im} \log \left [ \prod_{n=1}^\infty \left ( 1 + \frac{i}{(n+1)(n^2+1)} \right ) \right ] 
\end{align*}
I tried to combine it with the famous Euler product
$$ \frac{\sin \pi z}{\pi z} = \prod_{n=1}^{\infty} \left( 1 - \frac{z^2}{n^2} \right) \tag{1} $$
but I see no connection. So, is there a possible way to evaluate it?
 [/INST] We have
$$S = \sum_{n=1}^{\infty}\arctan\left(\frac{1}{n^3 + n^2 + n + 1}\right) \\ = \Im\left(\ln\left(\prod_{n=1}^{\infty} \left(1 + \frac{i}{n^3+n^2+n+1}\right)\right)\right) \\ = \Im\left(\ln\left(\prod_{n=1}^{\infty} \frac{n^3+n^2+n+1+i}{n^3+n^2+n+1}\right)\right) \\ = \Im\left(\ln\left(\frac{\pi\operatorname{csch}(\pi)}{1+i}\prod_{k=1}^{3}\frac{1}{\Gamma(r_k)}\right)\right)$$
where $r_k$ is the $k$th root of $x^3-x^2+x-1-i$ (order doesn't matter since the product iterates over all of them). The last step comes from employing equation 19 on here (which comes directly from the Weierstrass factorization product formula for $\Gamma(x)$), and the fact that $\Gamma(i)\Gamma(-i) = \pi\operatorname{csch}(\pi)$.
</s>