<s>[INST] Convergence of the series: $\sum_{n=2}^{\infty}\frac{1}{n^2(\ln n)^2\left | \sin(n\pi\sqrt{2}) \right |}$ 
Does the series $\sum_{n=2}^{\infty}\frac{1}{n^2(\ln n)^2\left | \sin(n\pi\sqrt{2}) \right |}$ converge or diverge?

Could you give me some hints? Thanks for helping.
 [/INST] Easier than it looks. Let $m$ be the closest integer to $n \sqrt 2.$ We have
$$  | 2 n^2 - m^2 | \geq 1. $$ 
they are integers and $\sqrt 2$ is irrational. $2 n^2 - m^2 \neq 0$ is an integer.
$$  | n \pi \sqrt 2 - m \pi  | = \left| \frac{\pi (2 n^2 - m^2)}{n \sqrt 2 + m} \right| \geq  \frac{\pi }{n \sqrt 2 + m} > \frac{\pi}{3n\sqrt 2} > \frac{1}{2n}  $$
This is as $n$ gets large... Therefore
$$ \left| \sin \left(n \pi \sqrt 2 \right) \right| > \frac{1}{3n} $$
and the thing you are summing is smaller than
$$ 3 \; \; \left( \frac{1}{n \log^2 n} \right) $$
which sum converges by the integral test
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