Does $\sum \frac{1}{n^x}$ converge uniformly?

Question

$\sum \frac{1}{n^x}$ for $x \in \left[\frac{4}{\pi},\infty\right)$ converge uniformly.

I'm trying to use the Weierstrass M-Test but having trouble finding an $M_n$.

Any hints?

Answer 1

Let $\sum f_n$ denote the series. Note that this series converges uniformly on $[A, \infty)$ for all $A > 1$: we have for all $x \ge A$ and all $n$, $n^x \ge n^A$, hence $1/n^x \le 1/n^A$. Thus,

$$\sup_{x \ge A} f_n(x) \le \frac1{n^A}$$

So you can use Weierstrass' M-test.

Note

The series does not converge uniformly on, say, $(1, \infty)$:

$$R_n(x) = \sum_{k=0}^{\infty} f_k (x) - \sum_{k=0}^n f_k(x) = \sum_{k=n+1}^{\infty} \frac1{k^x} > \sum_{k=n+1}^{2n} \frac1{k^x} > \frac{n}{(2n)^x} = \frac1{2 (2n)^{x-1}}$$

Putting $x_n = 1 + 1/n$, we get:

$$R_n(x_n) \ge \frac1{2(2n)^{1/n}} \to 1/2$$

Hence,

$$\sup_{x > 1} R_n(x) \not \to 0$$