Show that $\sum_{n=1}^{\infty} \frac{1}{n}\sin\bigg( \frac{x^n}{\sqrt{n}} \bigg)$ converges uniformly on the interval $[-1, 1]$ 
Show that
$$\sum_{n=1}^{\infty} \frac{1}{n}\sin\bigg( \frac{x^n}{\sqrt{n}} \bigg)$$
converges uniformly on the interval $[-1, 1]$
Hint: $|\sin(x)|\leq |x|$ for all $x\in \mathbb{R}$.

First step must be to show that the sum is convergent.
Using the hint, I assume I'm supposed to set up the inequality $ \bigg| \frac{1}{n}\sin\bigg( \frac{x^n}{\sqrt{n}} \bigg) \bigg| \leq \bigg| \frac{x^n}{n} \bigg|$.
I checked with Maple and found that $\sum_{n=1}^{\infty} \bigg| \frac{x^n}{n} \bigg| \approx 4.605\:$ by setting $x=0.99$ which suggests that both $\sum_{n=1}^{\infty} \bigg| \frac{x^n}{n} \bigg|$ and $\sum_{n=1}^{\infty} \frac{1}{n}\sin\bigg( \frac{x^n}{\sqrt{n}} \bigg)$ are convergent when $|x|<1$. But I'm not sure if this is true.
Furthermore, I'm supposed to show that the sum converges uniformly even if $x=1$. I'm not sure how to proceed.
 A: 
Show that
$$\sum_{n=1}^{\infty} \frac{1}{n}\sin\bigg( \frac{x^n}{\sqrt{n}} \bigg)$$
converges uniformly on the interval $[-1, 1]$

Use the given hint: $\left|\sin\left( \frac{x^n}{\sqrt{n}}\right)\right|\le \left|\frac{x^n}{\sqrt{n}}\right|\le \frac{1}{\sqrt{n}}$
So the series:
$$\sum_{n=1}^{\infty} \left|\frac{1}{n}\sin\bigg( \frac{x^n}{\sqrt{n}} \bigg)\right|\le \sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$$
Now you can use Weierstrass M-test.
A: Take the series for  $|\frac{x^{n}}{n^{\frac{3}{2}}}|$ This is the inequality since you should multiply what is in the sine by $\frac{1}{n}$.
So if we set x = 1 or -1(absolute value so doesnt matter) then by the p-test it converges since $\frac{3}{2}$ > 1.
If |x| < 1 then we have $|\frac{x^{n}}{n^{\frac{3}{2}}}|$ < $|\frac{1}{n^{\frac{3}{2}}}|$. There is a theorem(I forgot the name) that if $b_{n} < a_{n}$ and the sum of $a_{n}$ converges then so does the sum of $b_{n}$. So the series for  $|\frac{x^{n}}{n^{\frac{3}{2}}}|$ converges  overall for [-1,1] for x.
By that theorem $\sum_{n=1}^{\infty} |\frac{1}{n}\sin\bigg( \frac{x^n}{\sqrt{n}} \bigg)|$ < $\sum_{n=1}^{\infty} |\frac{x^{n}}{n^{\frac{3}{2}}}|$, so $\sum_{n=1}^{\infty} |\frac{1}{n}\sin\bigg( \frac{x^n}{\sqrt{n}} \bigg)|$ converges
There is another then that if the series for $|a_{n}|$ converges then so does the series for $a_{n}$. So  $\sum_{n=1}^{\infty} \frac{1}{n}\sin\bigg( \frac{x^n}{\sqrt{n}} \bigg)$ converges
