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

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?

• I would be surprised if the sum would have a closed form. Are you content with a numerical result ? – Peter Mar 1 at 12:33
• Can't I have any closed form? – Arjun Banerjee Mar 1 at 12:35
• wolfram gives only an approximate value: sum_(n=1)^∞ sin(1/n)/n = 1.47283 – Mitjackson Mar 1 at 12:36
• Where does this show up? Why do you need / do you believe there is a closed form? – punctured dusk Mar 1 at 12:44
• To prove the sum exists, that is a good question for a real analysis test. But to evaluate the sum, that would be very surpising for a real analysis test. – GEdgar Mar 1 at 14:06

\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.