Convergence of Series $sum_{n=1}^{infty} left[ frac{sin left( frac{n^2+1}{n}xright)}{sqrt{n}}left( 1+frac{1}{n}right)^nright]$

Consider the series $\displaystyle{ \sum_{n=1}^{\infty} \left[
\frac{\sin \left( \frac{n^2+1}{n}x\right)}{\sqrt{n}}\left(
1+\frac{1}{n}\right)^n\right]}$ . Find all points at which the series
is convergent. Find all intervals that the series is uniformly
convergent.

I know that I need to use Dirichlet and/or Abel Criterion to show this.

My first attempt was to consider the argument of the sum as product of three sequences of functions, $f_n,g_n,h_n$ and perform Dirichlet/Abel twice. Towards that end I tried to break up
$ \sin \left( \frac{n^2+1}{n}x\right)$ using $\sin(\alpha +\beta)=\sin \alpha \cos \beta \,+\, \sin \beta \cos \alpha $ to produce

$\sin \left(nx +\frac{x}{n} \right)=\sin(nx)\cos\left(\frac{x}{n}\right)+\sin\left(\frac{x}{n}\right)\cos (nx)$ and use the fact that $\sin(nx)\leq \frac{1}{\big|\sin\left(\frac{x}{2}\right)\big|}$, but then I don't know what to do with the $\sin\left(\frac{x}{n}\right)$, $\cos\left(\frac{x}{n}\right)$, and $\cos(nx)$. Previously, when using this method we showed uniform convergence on compact intervals like $[2k\pi-\varepsilon,2(k+1)\pi-\varepsilon]$.

Can I just say $ \frac{\sin \left( \frac{n^2+1}{n}x\right)}{\sqrt{n}}\leq \frac{1}{\sqrt{n}}\to 0$ as $n\to \infty$ ?

Also, I know that $\lim_{n\to \infty}\left( 1+\frac{1}{n}\right)^n=e$.

i.e. Help :) Thank you in advance

Answer

Note that

$$\sin \left( \frac{n^2+1}{n}x\right)=\sin \left( nx+\frac{1}{n}x\right)=\sin(nx)\cos\left(\frac{1}{n}x\right)+\cos(nx)\sin\left(\frac{1}{n}x\right)=$$

$$=\sin(nx)+\frac{x\cos(nx)}{n}+o\left(\frac1n\right)$$

$$\left( 1+\frac{1}{n}\right)^n=e^{n\log\left( 1+\frac{1}{n}\right)}=e^{1+\frac1n+o\left(\frac1n\right)}=e\left({1+\frac1n+o\left(\frac1n\right)}\right)$$

thus

$$\displaystyle{ \sum_{n=1}^{\infty} \left[
\frac{\sin \left( \frac{n^2+1}{n}x\right)}{\sqrt{n}}\left(
1+\frac{1}{n}\right)^n\right]}= e\sum_{n=1}^{\infty}\left[\frac{\sin(nx)}{\sqrt n}+\frac{x\cos(nx)}{n\sqrt n}+o\left(\frac1{n\sqrt n}\right)\right]$$

which converges since

$\sum_{n=1}^{\infty}\frac{\sin(nx)}{\sqrt n}$ converges by Dirichlet's test

$\sum_{n=1}^{\infty}\frac{x\cos(nx)}{n\sqrt n}$ converges absolutely by comparison with $\sum_{n=1}^{\infty}\frac{1}{n\sqrt n}$