# 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.