Convergence of series $\sum_{n=1}^\infty\sin(n)\cdot\sin(1/n)$

Question

I was doing math and I came across this series

$$\sum_{n=1}^\infty\sin n\cdot\sin\left(\dfrac1n\right)$$

I managed to prove that this series is convergent, but I was wondering what value the series converges to $?$ I've put it on Scientific workplace and got a value of around $0.92 $ But I have no idea how to get there.

Answer 1

The limit should be $0$.

An easy way to think about it is when n gets large, $\sin(\frac{1}{n})\approx \frac{1}{n}$. Then, we have $\lim_{n\to\infty}\sin(n)\sin(\frac{1}{n})=\lim_{n\to\infty}\frac{\sin(n)}{n}=0$ because $\sin(n)$ is bounded and $n$ is unbounded.

Answer 2

We have $$ 0 \leq |\sin n \cdot \sin \frac1n| = |\sin n| \cdot |\sin \frac1n| \leq |\sin \frac1n| \to |\sin 0| = |0| = 0. $$

Thus, $$\sin n \cdot \sin\frac1n \to 0.$$

Answer 3

You have a product of 2 sequences. The first is bounded and the second has limit 0. So the product should be a sequence with limit 0.