Infinite trigonometry Summation: $ \sum_{k=1}^\infty \left( \cos \frac{\pi}{2k}-\cos \frac{\pi}{2(k+2)} \right) $ I would like to evaluate
$$
\sum_{k=1}^\infty \left( \cos \frac{\pi}{2k}-\cos \frac{\pi}{2(k+2)} \right)
$$
Summation image please view before solving
I saw a pattern and realized the answer will converge and the final summation will be 1-(1/√2) but I am going wrong somewhere .Answer given is 2-(1/√2)
Plz help

This may be seen as a telescoping series, one may write for $N\ge1$,
$$
\small{\sum_{k=1}^N\! \left( \cos \frac{\pi}{2k}-\cos \frac{\pi}{2(k+2)} \right)\!=\sum_{k=1}^N \!\left(\! \cos \frac{\pi}{2k}-\cos \frac{\pi}{2(k+1)}\! \right)\!+\!\sum_{k=1}^N\! \left(\! \cos \frac{\pi}{2(k+1)}-\cos \frac{\pi}{2(k+2)} \!\right)}
$$ giving
$$
\small{\sum_{k=1}^N \! \left( \cos \frac{\pi}{2k}-\cos \frac{\pi}{2(k+2)} \right)=\!\!\left(\cos \frac{\pi}{2}-\cos \frac{\pi}{2(N+1)}\right)\!+\!\left(\cos \frac{\pi}{4}-\cos \frac{\pi}{2(N+2)}\right)}
$$ then by letting $N \to \infty$, one gets
$$
\sum_{k=1}^\infty \left( \cos \frac{\pi}{2k}-\cos \frac{\pi}{2(k+2)} \right)=(0-1)+\left(\cos \frac{\pi}{4}-1\right)
$$I think you can take it from here.
