Good morning,
recently in a school near mine there was a contest for solving a infinite series that even professors couldn't, only thing we knew was the final result, approximately 0.118272222745.
The infinite series was this:
$$ \sum_{n=1}^{\infty} \frac{\cos(n)}{n(n+1)} $$
I'm trying to find a closed-form expression for this series, if it exists. Has anyone come across a series like this or have any insights on how to approach it and eventually solve it?
Best Answer
$$\eqalign{ & \sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{{n\left( {n + 1} \right)}} = \frac{{\ln \left( {1 - x} \right)}}{x} - \ln \left( {1 - x} \right) + 1} \cr & \Rightarrow \sum\limits_{n = 1}^\infty {\frac{{\cos \left( n \right)}}{{n\left( {n + 1} \right)}} = \Re \sum\limits_{n = 1}^\infty {\frac{{{e^{in}}}}{{n\left( {n + 1} \right)}} = \Re } } \left( {\frac{{\ln \left( {1 - {e^i}} \right)}}{{{e^i}}} - \ln \left( {1 - {e^i}} \right) + 1} \right) \cr & = 1 + \frac{1}{2}\sin \left( 1 \right) - \frac{1}{2}\pi \sin \left( 1 \right) - \frac{1}{2}\ln \left( {2 - 2\cos \left( 1 \right)} \right) + \frac{1}{2}\cos \left( 1 \right)\ln \left( {2 - 2\cos \left( 1 \right)} \right) \cr & = 1 + \frac{1}{2}\sin \left( 1 \right)\left( {1 - \pi } \right) - {\sin ^2}\left( {\frac{1}{2}} \right)\ln \left( {2 - 2\cos \left( 1 \right)} \right) \cr} $$