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

Question

$$ \sum_{10}^{\infty} \frac{\sin{\frac{1}{n}}}{\ln(n)}\left(e^{\frac{1}{n^2}} - 1\right)\left(\sqrt{n^4 - 8}\right) $$

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.

Answer 1

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 $(7)$.

Answer 2

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.

Answer 3

For large $n$, your sequence behaves as $$\frac{1}{n\log n}$$ Because the following integral diverges, $$\int_{10}^\infty \frac{dx}{x\log x}=\log \log x \Big|_{10}^\infty $$ By the integral test, your series also diverges.