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?

  • 6
    $\begingroup$ I would be surprised if the sum would have a closed form. Are you content with a numerical result ? $\endgroup$ – Peter Mar 1 at 12:33
  • $\begingroup$ Can't I have any closed form? $\endgroup$ – Arjun Banerjee Mar 1 at 12:35
  • $\begingroup$ wolfram gives only an approximate value: sum_(n=1)^∞ sin(1/n)/n = 1.47283 $\endgroup$ – Mitjackson Mar 1 at 12:36
  • 1
    $\begingroup$ Where does this show up? Why do you need / do you believe there is a closed form? $\endgroup$ – punctured dusk Mar 1 at 12:44
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    $\begingroup$ 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. $\endgroup$ – GEdgar Mar 1 at 14:06

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.

  • 1
    $\begingroup$ The Taylor expansion of the sine function is wrong. $\endgroup$ – Marco Cantarini Mar 2 at 14:16
  • $\begingroup$ I have fixed the problem. Thank you. $\endgroup$ – kelvin hong 方 Mar 3 at 5:03

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