Why $ \sum_{k=1}^{\infty}\frac{1}{k^{1+x}} $ is not uniformly convergent on $ \left(0,\;+\infty\right) $

Question

Why this series of functions is not uniformly convergent on $ \left(0,\;+\infty\right) $. \begin{equation} \sum_{k=1}^{\infty}\frac{1}{k^{1+x}} \end{equation}

Answer 1

Hint: What do you think happens for $x\to 0^+$?

Answer 2

This is the famous Riemann $\zeta$ function, shifted by 1 to the left: https://en.wikipedia.org/wiki/Riemann_zeta_function

It is well-known that is has a singularity at 1 (in your notation due to the shift, at 0), and it is a simple pole. In other words, around 1, the $\zeta$ function around 1 "looks like" $1/(x-1)$. To make it more precise, $\zeta(x)-1/(x-1)$ is a smooth function on the positive half of the complex plane.

To translate all of this to your function: it has a simple pole at 0, and around 0, it looks like $1/x$, which is not uniformly continuous.

Answer 3

Hint:

$$\sup_{x \in (0,\infty)}\sum_{k=n+1}^{\infty} \frac{1}{k^{1+x}} >\sum_{k=n+1}^{2n} \frac{1}{k^{1+1/n}} > \frac{n}{(2n)^{1+1/n}}$$

Answer 4

Uniform convergence means that $$\sup_{x\in(0, +\infty)} \left|\sum_{k=1}^\infty \frac1{k^{1+x}} - \sum_{k=1}^n\frac1{k^{1+x}}\right| \xrightarrow{n\to\infty} 0$$

As @RRL suggests, for $n\in\mathbb{N}$ you can consider $x_n = \frac1n$ so

\begin{align} \sup_{x\in(0, +\infty)} \left|\sum_{k=1}^\infty \frac1{k^{1+x}} - \sum_{k=1}^n\frac1{k^{1+x}}\right| &= \sup_{x\in(0, +\infty)} \sum_{k=n+1}^\infty\frac1{k^{1+x}} \\ &\ge \sum_{k=n+1}^\infty\frac1{k^{1+\frac1n}}\\ &\ge \int_{n+1}^\infty \frac{dt}{t^{1+\frac1n}}\\ &= -\frac{n}{t^{1/n}} \Bigg|_{n+1}^\infty\\ &= \frac{n}{\sqrt[n]{n+1}}\\ &\xrightarrow{n\to\infty} +\infty \end{align}

so it doesn't converge to $0$.