TITLE: Convergence of sequences $\sum_{k=1}^n \frac{1}{k}\sin\frac{n}{k}$ as $n\to\infty$.
QUESTION [7 upvotes]: My question is to prove the limit
$$\lim_{n\to\infty}\sum_{k=1}^n \frac{1}{k}\sin\frac{n}{k}$$
does not exist!
$\textbf{Background:}$
Generally, we have the following result:

If $f$ is monotone on $(0,1]$, then
$$\int_0^1 f(x) \,dx\ \mbox{exists} \iff \lim_{n\to\infty} \frac1{n}\sum_{k=1}^nf\left(\frac{k}{n}\right)\ \mbox{exists}.$$
And in both case, we have
$$\int_0^1 f(x) \,dx=\lim_{n\to\infty} \frac1{n}\sum_{k=1}^nf\left(\frac{k}{n}\right).$$

If $f$ has no monotonicity, the above result may not be right.For example, let
$$f(x)=\frac{1}{x}\sin\frac{1}{x},\ x\in(0,1].$$
It easy to see that
$$\int_{0}^{1}\frac{1}{x}\sin\frac{1}{x}dx
=\int_{1}^{\infty}\frac{\sin u}{u}du=0.624713\cdots.$$
But the limit
$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n \frac{n}{k}\sin\frac{n}{k}=\lim_{n\to\infty}\sum_{k=1}^n \frac{1}{k}\sin\frac{n}{k}$$
doesn't seem to exist by numerical calculation using wolfram mathematica:For $n=1,2,\cdots,200$

Can someone provide a method to prove above limit does not exist.
Any help and hints will welcome!

REPLY [6 votes]: Let $a_n=\sum_{k=1}^n \frac{1}{k}\sin\frac{n}{k}$. To show that $a_n$ doesn't converge, it is enough to show that $a_{n+1}-a_n$ doesn't go to zero.
But $a_{n+1}-a_n=2\sin 0.5\cos \frac{2n+1}{2}+\sum_{k=2}^n2\frac{1}{k}\sin \frac{1}{2k}\cos \frac{2n+1}{2k}+ \frac{1}{n+1}\sin 1$
Now $2\sin 0.5=.988..$, while $|\sum_{k=2}^n2\frac{1}{k}\sin \frac{1}{2k}\cos \frac{2n+1}{2k}| \le \sum_{k=2}^{\infty}\frac{1}{k^2} \le .65$ and $\frac{1}{n+1}\sin 1 \to 0$, so it is enough to find $n_m \to \infty$ st $\cos \frac{2n_m+1}{2}$ is close to $\pm 1$ for $|a_{n_m+1}-a_{n_m}| \ge .1$ say and by the uniform distribution of $\frac{2n+1}{2}$ modulo $2\pi$ we can definitely find such, so we are done!