why is $\sum_{n=1}^{\infty} \frac{1}{n} \sin\left( \frac{x^n}{\sqrt n} \right)$ uniformly convergent? I need a argument for why the $\sum_{n=1}^{\infty}  \frac{1}{n} \sin\left(\frac{x^n}{\sqrt n}\right)$ is uniformly convergent for $x = [-1,1]$
I know that the $\sin\left( \frac{x^n}{\sqrt n} \right)$ is convergent, but how do I know that the sum is convergent?
The sum $\sum_{n=1}^{\infty}\frac{1}{n}$ isn't convergent, so how do I know that the product is convergent?
 A: Use Weirestrass M test. We have $|\sin(x)|\leq |x|,\,\forall x\in\mathbb{R}$
$\displaystyle|\frac{1}{n} \sin\left(\frac{x^n}{\sqrt n}\right)|\leq|\frac{1}{n}||\frac{x^{n}}{\sqrt{n}}|\leq\frac{1}{n}\frac{1}{\sqrt{n}}$. This is because $x\in[-1,1]$
Since $\sum_{n=1}^{\infty}\frac{1}{n\sqrt{n}}$ converges you have uniform convergence by Weirestrass-M Test .
A: It's a serie of definitely alternating sign. So, we must study the absolute convergence.
Let:
$$a_n=\left|\frac{1}{n} \sin\left(\frac{x^n}{\sqrt n}\right)\right|$$
We can apply the asympotic criteria because $a_n$ is always non-negative. We have:
$$\lim_{n\to+\infty}a_n=\lim_{n\to +\infty}\left|\frac{1}{n} \sin\left(\frac{x^n}{\sqrt n} \right)\right| \sim\lim_{n\to+\infty}\left|\frac{x^n}{n\cdot \sqrt n}\right| \leq\frac{1}{n\cdot \sqrt n}=\frac{1}{n^{\frac{3}{2}}}$$
Here, we have use the fact that $\frac{|x^n|}{\sqrt{n}}\to 0$ if $x\in [-1,1]$. The series $\sum_{n=1}^{+\infty}\frac{1}{n^{\frac{3}{2}}}$ converges, thus by comparison test, converges also your series.
