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. 

Because $\sin(x)$ is concave on $\left[0,\frac\pi2\right]$, for $x\in\left[0,\frac\pi2\right]$,
$$
\frac{\sin(x)-\sin(0)}{x-0}\ge\frac{\sin\left(\frac\pi2\right)-\sin(0)}{\frac\pi2-0}\tag1
$$
Therefore,
$$
\sin(x)\ge\frac{2x}\pi\tag2
$$
Since $e^x$ is convex,
$$
\begin{align}
\frac{e^x-e^0}{x-0}
&\ge\lim_{t\to0}\frac{e^t-e^0}{t-0}\\
&=e^0\\[6pt]
&=1\tag3
\end{align}
$$
Therefore,
$$
e^x\ge1+x\tag4
$$
and that
$$
\begin{align}
x-\sqrt{x^2-a}
&=\frac{a}{x+\sqrt{x^2-a}}\\
&\le\frac ax\tag5
\end{align}
$$
Therefore,
$$
\sqrt{x^2-a}\ge x-\frac ax\tag6
$$
Thus, applying $\color{#C00}{(2)}$, $\color{#090}{(4)}$, and $\color{#00F}{(6)}$, for $n\ge2$, we have
$$
\begin{align}
\frac{\color{#C00}{\sin\left(\frac1n\right)}}{\log(n)}\color{#090}{\left(e^{\frac1{n^2}}-1\right)}\color{#00F}{\sqrt{n^4-8}}
&\ge\frac{\color{#C00}{\frac2{\pi n}}}{\log(n)}\color{#090}{\left(\frac1{n^2}\right)}\color{#00F}{\left(n^2-\frac8{n^2}\right)}\\
&=\frac2\pi\frac1{n\log(n)}\left(1-\frac8{n^4}\right)\\
&\ge\frac1\pi\frac1{n\log(n)}\tag7
\end{align}
$$
The series $\sum\limits_{n=2}^\infty\frac1{n\log(n)}$ diverges by the Cauchy Condensation test (applied twice is nice); and therefore, the original series diverges by comparison with $\sum\limits_{n=2}^\infty\frac1{n\log(n)}$ using $(7)$.
