On proving uniform convergence on an interval. I have a doubt on uniform convergence. We have said in class that a series of functions $f_n(x)$ converges uniformly on an interval $I$ iff 
$$ \lim_{n \rightarrow \infty} \sup_{I} | \sum_{k = n +1}^{n+p} f_n(x) | = 0 \quad \quad (1)  $$
But we also said in class that the series of functions given by $$\sum_{n=1}^{\infty} \frac{1}{n^x}$$ converges uniformly on any interval of the form $[c, + \infty)$ with $c > 1.$ But does not converge uniformly on the interval $(1, + \infty)$.
But if I try to check with $(1)$ I obtain $$ \sup_{(1, + \infty)} | \sum_{k = n +1}^{n+p} \frac{1}{n^x} | = \sup_{(1, + \infty)} |\frac{1}{n^x}+ \frac{1}{(n+1)^x} + \dots + \frac{1}{(n+p)^x} | = \frac{1}{n}+ \frac{1}{(n+1)} + \dots + \frac{1}{(n+p)} $$
and taking the limit of this gives me zero, so it should converge uniformly even on $(1, \infty)$. This means I have made some kind of mistake on my notes, anybody mind telling me where?
 A: $u_n\to u$ uniformly over $I$ iff
$$\lim_{n\to\infty}\sup_{x\in I}|u_n(x)-u(x)|=0
$$
Now let $u_n(x)=\sum_{k=1}^nf_k(x)$ and $u(x)=\sum_{k=1}^\infty f_k(x)$. So you have to replace the upper limit $n+p$ with $\infty$ to get the correct condition for uniform convergence and note that the last sum you've listed becomes the tail of harmonic series which is well known to be divergent.
A: To prove that the series is not uniformly convergent on $(1,+\infty)$ you now can make use of a sum with finite limits.
Negating the definition, the series $\sum_{k=1}^{\infty}f_k(x)$ fails to converge uniformly on an interval $I$ if there exists some $\epsilon_0 > 0$ such that for every $n \in \mathbb{N}$ we have $\sup_{x \in I}|\sum_{k=n+1}^{\infty}f_k(x)| \geqslant \epsilon_0.$ It is not necessary for the supremum of the tail to go to infinity -- it just has to not converge to $0$ to show non-uniform convergence.
In this case we have for all $n$
$$\sup_{x \in (1,\infty)}\left|\sum_{k=n+1}^{\infty}\frac{1}{k^x}\right| \geqslant \sup_{x \in (1,\infty)}\left|\sum_{k=n+1}^{2n}\frac{1}{k^x}\right| \geqslant \sup_{x \in (1,\infty)}\frac{n}{(2n)^x} = \frac{n}{2n}= \frac1{2}.$$
