# Showing $\sum_{n=1}^\infty n^{-x}$ *doesn't* converge uniformly for $x \in (1,\infty)$?

A question says, show that $\sum_{n=1}^\infty n^{-x}$ converges pointwise but not uniformly for $x \in (1,\infty)$. I can show it converges pointwise by taking $x\in (1+\delta, \infty)$ for any $x$ and $\delta>0$ and then using the Weierstrass-M test on $1+\delta$.

But I'm struggeling to show that it doesn't converge uniformly? Thanks!

Hint: The series diverges to infinity as $x\rightarrow 1$. In particular, for any $m$, we cannot bound $$\sum_{n=m}^\infty n^{-x}$$ uniformily, that is independently of $x$, for $x>1$.

• Could you elaborate? I can't see how to show this directly?
– user26069
Apr 22, 2012 at 13:02
• @user26069: Tell me the exact definition of uniform convergence, and we can proceed from there. Apr 22, 2012 at 13:18
• $s_n \to a$ uniformly if $\forall \epsilon > 0, \exists N = N(\epsilon)$ such that $(n \geq N$ and $x \in (1,\infty)) \implies |s_n - a| < \epsilon$
– user26069
Apr 22, 2012 at 13:41
• Where $s_n = \sum_{r=1}^n r^{-x}$
– user26069
Apr 22, 2012 at 13:42
• Is that right? Where do we go from here?
– user26069
Apr 22, 2012 at 14:18