Determine whether $\sum\limits_{n=1}^\infty \frac{1}{n^x}$ converges uniformly on $(1,\infty)$

Question

Detemine whether $\sum\limits_{n=1}^\infty \frac{1}{n^x}$ converges uniformly on $(1,\infty)$.

My attempt: Upon attempting to use the Weierstrauss M-test I get $$0\leqslant\|f_n(x)\|_\infty=\sup_{x\in (1,\infty)}|\frac{1}{n^x}|\leqslant\frac{1}{n}=M_n$$ But by definition, $\sum\limits_{n=1}^\infty M_n=\sum\limits_{n=1}^\infty \frac{1}{n}$ diverges. So the Weierstrauss M-test is not useful here. Is there some way I could possibly use the uniform Cauchy principle? Thanks for the help.

Answer 1

Idea. If it converges uniformly then $\sum_{n=1}^\infty\frac{1}{n}$ have to converge.

Hint

Suppose it converges uniformly. Let's take any $\varepsilon > 0$. Then there is some $N$ such that $\forall x \in (1, \infty):\sum_{n=N}^\infty\frac{1}{n^x}<\varepsilon$. Let's find $M > N$ such that $\sum_{n=N}^M\frac{1}{n}>2\varepsilon.$

Final shot

It is obvious that if $x \rightarrow 1$ then $\sum_{n=N}^M\frac{1}{n^x}\rightarrow \sum_{n=N}^M\frac{1}{n} > 2\varepsilon$ (since everything is finite). But how can it be?