<s>[INST] how to show $ \sum_{10}^{\infty} \frac{\sin{\frac{1}{n}}}{\ln(n)}(e^{\frac{1}{n^2}} - 1)(\sqrt{n^4 - 8})$ converge/diverge $$ \sum_{10}^{\infty} \frac{\sin{\frac{1}{n}}}{\ln(n)}\left(e^{\frac{1}{n^2}} - 1\right)\left(\sqrt{n^4 - 8}\right) $$
I have tried a lot of stuff, didn't work at all . A hint will be good too. 
I know that $\sin(\frac{1}{n}) < \frac{1}{n}$. I have tried to show with Cauchy that it diverges. 
 [/INST] Hint. One may observe that, 
$$
\lim_{ n\to \infty}(e^{\frac{1}{n^2}} - 1)(\sqrt{n^4 - 8})= 1 
$$giving, for a certain $n_0\ge10$, $n\ge n_0$,
$$
\frac12 \le(e^{\frac{1}{n^2}} - 1)(\sqrt{n^4 - 8}) \tag1
$$ then, as $n \ge 10$,
$$
\frac{1}{n}-\frac{1}{6n^3}\le\sin{\frac{1}{n}}\tag2
$$ yielding, for $N\ge n_0$,
$$
\frac{1}{2}\sum_{n_0}^{N}\frac{1}{n\ln n}-\frac{1}{12}\sum_{n_0}^{N}\frac{1}{n^3\ln n}\le\sum_{n_0}^{N} \frac{\sin{\frac{1}{n}}}{\ln n}(e^{\frac{1}{n^2}} - 1)(\sqrt{n^4 - 8})
$$ leading to the divergence of the given series.
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