Find $\lim_{n\to \infty}\sum_{k=1}^{n}(\sin\frac{\pi}{2k}-\cos\frac{\pi}{2k}-\sin\frac{\pi}{2(k+2)}+\cos\frac{\pi}{2(k+2)})$

Question

Find $\lim_{n\to \infty}\sum_{k=1}^{n}\left(\sin\frac{\pi}{2k}-\cos\frac{\pi}{2k}-\sin\frac{\pi}{2(k+2)}+\cos\frac{\pi}{2(k+2)}\right)$

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$$\lim_{n\to \infty}\sum_{k=1}^{n}\left(\sin\frac{\pi}{2k}-\cos\frac{\pi}{2k}-\sin\frac{\pi}{2(k+2)}+\cos\frac{\pi}{2(k+2)}\right)$$ $$\lim_{n\to \infty}\sum_{k=1}^{n}\left(\sin\frac{\pi}{2k}-\sin\frac{\pi}{2(k+2)}-\cos\frac{\pi}{2k}+\cos\frac{\pi}{2(k+2)}\right)$$ $$\lim_{n\to \infty}\sum_{k=1}^{n}\left(2\cos\frac{(\frac{\pi}{2k}+\frac{\pi}{2(k+2)})}{2}\sin\frac{(\frac{\pi}{2k}-\frac{\pi}{2(k+2)})}{2}+2\sin\frac{(\frac{\pi}{2k}+\frac{\pi}{2(k+2)})}{2}\sin\frac{(\frac{\pi}{2k}-\frac{\pi}{2(k+2)})}{2}\right)$$ I am stuck here.

Answer 1

Hint. One may recall teslescoping sums, by writing $$ u_{k+2}-u_k=\left(u_{k+2}-u_{k+1} \right)+\left(u_{k+1}-u_{k} \right) $$ giving $$ \sum_{k=1}^n\left(u_{k+2}-u_{k} \right)=\left(u_{n+2}-u_2\right)+\left(u_{n+1}-u_1\right). $$