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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!

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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 $x>1$.

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Could you elaborate? I can't see how to show this directly? – user26069 Apr 22 '12 at 13:02
@user26069: Tell me the exact definition of uniform convergence, and we can proceed from there. – Eric Naslund Apr 22 '12 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 '12 at 13:41
Where $s_n = \sum_{r=1}^n r^{-x}$ – user26069 Apr 22 '12 at 13:42
Is that right? Where do we go from here? – user26069 Apr 22 '12 at 14:18

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