How to prove that $\sum(-1)^n\sin \frac{1}{n^\alpha \ln n}$ is divergent for any $\alpha\leq 0$? 
How to prove that $\sum(-1)^n\sin \frac{1}{n^\alpha \ln n}$ is divergent for any $\alpha\leq 0$?

It is well-known that $\lim \sin n$ does not exist. But that procedure could not be easily adapted to those that $\lim \sin \frac{1}{n^\alpha\ln n}$ does not exists. How to do then for this above problem?

Let $a_n=\frac{1}{n^\alpha \ln n}, n \ge 2$
For $\alpha=0$ we have that $a_n$ decreases to $0$ as $n \to \infty$ hence $\sin a_n$ does so even from $n=2$ on ($a_2=1/\ln 2 < \pi/2$) so the series converges by the alternating test.
If $0 < \beta=-\alpha \le 1$ we let $f_\beta(x)=f(x)=\frac{x^{\beta}}{\pi \ln x}, x>3/2$
Then $f'(x)=(\beta-\frac{1}{\ln x})\frac{f(x)}{x}$ so $f'(x)$ (eventually) decreases to zero as $x \to \infty$ (for monotonicity we use that the leading term of $f''$ which dominates is negative as it easily seen, so $f'$ eventually starts decreasing) and also $xf'(x) \to \infty, x \to \infty$ since $f(x) \to \infty, x \to \infty$.
Hence by Fejer's theorem, we get that $f(n)$ is equidistributed modulo $1$ so $\sin a_n=\sin \pi f(n)$  cannot have limit zero as $n \to \infty$, so the series doesn't converge.
If now $\beta>1$ and $k\ge 1$ the unique positive integer for which $0< \beta -k \le 1$, one applies the same ud modulo $1$ criterion as above but for the $k+1$ derivative of $f$ - namely, $f^{(k+1)}(x)$ decreses eventually to zero at infinity and $xf^{(k+1)}(x) \to \infty$ which is an easy inductive exercise since the leading term of $f^{(k)}$ will be $cf_{\beta-k}$ and which is as in the previous case.
This shows that $f(n)$ is still ud modulo $1$ by the easy inductive generalization of Fejer's theorem (using Van der Corput difference criterion) in this case hence $\sin \pi f(n)$ cannot converge to zero.
