Convergence of Series $\sum_{n=1}^{\infty} \left[ \frac{\sin \left( \frac{n^2+1}{n}x\right)}{\sqrt{n}}\left( 1+\frac{1}{n}\right)^n\right]$ 
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
 A: 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 $\sum_{n=1}^{\infty}\frac{1}{n\sqrt n}$
A: Pointwise convergence: Fix $x.$ Our series equals
$$\tag 1 \sum_{n=1}^{\infty} \frac{\sin(nx+x/n)-\sin (nx)}{\sqrt n}(1+1/n)^n + \sum_{n=1}^{\infty} \frac{\sin (nx)}{\sqrt n}(1+1/n)^n.$$
Now by the MVT,
$$|\sin(nx+x/n)-\sin (nx)| = |(\cos c)(x/n)| \le |x/n|.$$
So in the first series in $(1),$ the sum of the absolute values is bounded above by
$$\sum_{n=1}^{\infty} \frac{|x/n|}{\sqrt n}(1+1/n)^n.$$
Because $(1+1/n)^n\le e,$ this series converges absolutely, hence converges.
In the second series in $(1),$ let's use $(1+1/n)^{1/n} = e+ O(1/n).$ (I'll leave this as an exercise.) Then this series equals
$$\sum_{n=1}^{\infty} \frac{\sin (nx)}{\sqrt n}e + \sum_{n=1}^{\infty}\sin (nx)\cdot O\left (\frac{1}{n^{3/2}}\right).$$
Here the the first series converges by Dirichlet's test, and the second series converges absolutely. Thus $(1)$ converges pointwise everywhere.
Uniform convergence: From the above, after dropping the constant $e,$ our series can be written
$$\tag 2 \sum_{n=1}^{\infty} \frac{\sin (nx)}{\sqrt n} + \sum f_n(x),$$
where $\sum f_n(x)$ converges uniformly on $[-r,r]$ for every $r>0.$ That's not true of the first series. This series is $2\pi$-periodic, so consider what happens on $[0,2\pi].$ Summing by parts (a la Dirichlet) shows that the series converges uniformly on $[a,2\pi-a]$ for all small $a>0.$
Suppose the convergence were uniform to some $f$ on $[0,2\pi].$ Then $f$ would be continuous and we would have
$$\int_0^{2\pi}|f|^2 = \lim_{N\to \infty} \int_0^{2\pi}\left |\sum_{n=1}^{N} \frac{\sin (nx)}{\sqrt n}\right |^2\, dx = \lim_{N\to \infty}\sum_{n=1}^{N} \frac{\pi}{(\sqrt n)^2} = \infty.$$
In the second $=$ we have unsed the orthogonality of the functions $\sin (nx).$ That's a contradiction, hence uniform convergence on $[0,2\pi]$  fails. Putting this all together shows the intervals of convergence for $(2)$ are $[2m\pi + a,2\pi(m+1)-a],m\in \mathbb Z,$ for any small $a>0.$
