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?

Something tells me this might be difficult to do by hand. Plugging in your product formula into Mathematica and then Simplifying, this is what comes out:

