Why $ \sum_{k=1}^{\infty}\frac{1}{k^{1+x}} $ is not uniformly convergent on $ \left(0,\;+\infty\right) $ 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}
 A: Hint: What do you think happens for $x\to 0^+$?
A: 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. 
A: Hint:
$$\sup_{x \in (0,\infty)}\sum_{k=n+1}^{\infty} \frac{1}{k^{1+x}} >\sum_{k=n+1}^{2n} \frac{1}{k^{1+1/n}} >  \frac{n}{(2n)^{1+1/n}}$$
A: 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 $0$.
