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Question Number 80332 by ~blr237~ last updated on 02/Feb/20

$$let\alpha \in \mathbb{R}and{a}_n=\underset{k=1}{\sum ^n}\frac{sin\left(k\alpha \right)}{n+k}$$$$Find\underset{n\to \mathrm{\infty }}{\lim }{a}_n$$

Commented by Rio Michael last updated on 03/Feb/20

$$a_n=\frac{\sin \alpha }{n}+\frac{\sin 2\alpha }{n+2}+\frac{\sin 3\alpha }{n+3}+...$$$$\underset{n\to \mathrm{\infty }}{\lim }{a}_n=\underset{n\to \mathrm{\infty }}{\lim }(\frac{\sin \alpha }{n}+\frac{\sin 2\alpha }{n+2}+\frac{\sin 3\alpha }{n+3}+...)$$$$\mathrm{For}\mathrm{small}\mathrm{angles}\sin \alpha =\alpha$$$$\Rightarrow \underset{n\to \mathrm{\infty }}{\lim }{a}_n=\underset{n\to \mathrm{\infty }}{\lim }(\frac{\alpha }{n}+\frac{\sin 2\alpha }{n+2}+...)=0$$

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