Evaluation of the sum $\sum\limits_{n=1}^{\infty}\frac1n\sin\frac1n$

Question

I am trying to evaluate the sum $\displaystyle\sum_{n=1}^{\infty}\dfrac1n\sin\dfrac1n$.

This was given in my real analysis test yesterday.

I have proved that the sum exists:

We know for any non-negative real $x$, $\sin x\le x$.

Hence $$\displaystyle\sum_{n=1}^{\infty}\dfrac1n\sin\dfrac1n\le \displaystyle\sum_{n=1}^{\infty}\dfrac1n\cdot\dfrac1n=\displaystyle\sum_{n=1}^{\infty}\dfrac1{n^2}=\dfrac{\pi^2}{6}$$

But how can I find the sum?

Answer 1

I cannot say there is no closed form, I just hope this gives you an idea.

\begin{align*}\sum_{n=1}^\infty \frac1n\sin\frac1n&=\sum_{n=1}^\infty \frac1n\bigg[\frac1n-\frac1{3!n^3}+\frac1{5!n^5}-\frac1{7!n^7}+\cdots\bigg]\\ &=\sum_{n=1}^\infty\bigg[\frac1{n^2}-\frac1{3!n^4}+\frac1{5!n^6}-\frac1{7!n^8}+\cdots\bigg]\\ &=\zeta(2)-\frac16\zeta(4)+\frac1{120}\zeta(6)-\frac1{5040}\zeta(8)+\cdots\end{align*}

When $k$ get large, $\zeta(k)$ will get closer and closer to $1$, I believe this gives a faster convergent to the sum.