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!
Best Answer
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 $$a_{n+1}-a_n=2\sin\frac12\cos\frac{2n+1}{2} +\sum_{k=2}^n\frac{2}{k}\sin\frac{1}{2k}\cos\frac{2n+1}{2k} +\frac{\sin1}{n+1}.$$ Now $2\sin\frac12=0.988\ldots$, while $$\left|\sum_{k=2}^n\frac{2}{k}\sin\frac{1}{2k}\cos\frac{2n+1}{2k}\right|\le\sum_{k=2}^{\infty}\frac{1}{k^2}\le 0.65$$ and $\frac{\sin1}{n+1}\to0$, so it is enough to find $n_m \to \infty$ such that $\cos\frac{2n_m+1}{2}$ is close to $\pm 1$ for $|a_{n_m+1}-a_{n_m}|\ge 0.1$ say and by the uniform distribution of $\frac{2n+1}{2}$ modulo $2\pi$ we can definitely find such, so we are done!