show that the series $ \sum^{\infty}_{n=1} \frac{z^n}{n^{3/2}}$ converges uniformly for $|z|< 1$

Question

show that the series $ \sum^{\infty}_{n=1} \frac{z^n}{n^{3/2}}$ converges uniformly for $|z|< 1$

I don't know how to begin this problem, can someone tell me how can i demonstrate this.

Answer 1

HINT: For all $|z| \le 1$:

$$\left|\sum_{n=N}^{\infty} \frac{z^n}{n^{3/2}} \right| \ \le \ \sum_{n=N}^{\infty} \frac{|z|^n}{n^{3/2}} \ \le \ \sum_{n=N}^{\infty} \frac{1}{n^{3/2}}.$$

What do we know about $\sum_{n=N}^{\infty} \frac{1}{n^a}$ for all constants $a$ at least say 5/4?

Answer 2

Hint:

Writing in the form $\sum^{\infty}_{n=0} \frac{z^n}{n^{3/2}}=\sum^{\infty}_{n=0} a_nz^n,$ we see that $a_n=\frac{1}{n^{3/2}}$.

The radius of convergence is given by $R=\frac{1}{\limsup |a_n|^{1/n}}$.

Then, for any $z$ with $|z|\leq r<R$, the series will converge uniformly.