Uniform convergence of sum $(1/n)^x$

Question

I'm trying to check where $\sum_{n=1}^{\infty}1/n^x$ uniformly converges on $(1,\infty)$. I want to do that with the Cauchy criteria. For every $\epsilon$>0 exists an $N>0$ so that for all $n>N$ and all natural $P$: $$|\sum_{n=1}^{P}1/n^x|<\epsilon, \quad \text{for all} \ x>1. $$ (a subsequence of a known convergent sum). So for all $n>N$ and for all natural $P$ and $x>1$: $$ |\sum_{k=n+1}^{n+P}1/k^x|\leq|\sum_{n=1}^{P}1/n^x|<\epsilon. $$ Thus $\sum_{n=1}^{\infty}1/n^x$ uniformly converges for every $x>1$. Now I know this is probably lacking/wrong, so I ask you guys to correct me please :) Thanks.(sorry for not putting proper spacing, I don't know how to do that yet).

Answer 1

A sequence of functions $(f_n)_{n=1}^\infty$ converges uniformly to a function $f$ on an interval $I$ means: $\forall\epsilon>0$, $\exists n_0\in\mathbb{N}$ such that $$|f_N(x)-f(x)|<\epsilon,\quad\forall N>n_0,\forall x\in I,$$ This is equivalent to say that $$\lim_{N\to\infty}\sup_{x\in I}|f_N(x)-f(x)|=0.$$ Now back to the question, note that $$\sup_{x\in(1,\infty)}\sum_{n=N+1}^\infty \frac{1}{n^x}\geq \sup_{x\in(1,\infty)}\sum_{n=N+1}^M \frac{1}{n^x}=\sum_{n=N+1}^M\frac{1}{n},\quad\forall M>N+1$$ it follows that $$\sup_{x\in(1,\infty)}\sum_{n=N+1}^\infty \frac{1}{n^x}\geq\lim_{M\to\infty}\sum_{n=N+1}^M\frac{1}{n}=+\infty.$$ Thus $$\lim_{N\to\infty}\sup_{x\in(1,+\infty)}\sum_{n=N+1}^\infty \frac{1}{n^x}=+\infty,$$ and therefore the series does not converge uniformly on $(1,+\infty)$.