TITLE: Convergence of $\sum_{n=1}^{\infty}\frac{1}{n}\sin\left(\frac{1}{n}\right)$
QUESTION [12 upvotes]: Showing convergence is easy, since for $x\geq 0$ we have $\sin x\leq x$ so:
$$0\leq\sum_{n=1}^{\infty}\frac{1}{n}\sin\left(\frac{1}{n}\right)\leq \sum_{n=1}^{\infty}\frac{1}{n^2}$$
I was wondering if it might have a closed form? Finding one is beyond me, but I am sure that the people on here will have ideas.

REPLY [4 votes]: The sum is approximately $1.47282823195618529629494738$. Plugging this into the Inverse Symbolic Calculator fails to find a match. Most likely, this has no closed form.