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A question in a past paper says prove that this series converges pointwise but not uniformly $$\xi(x):= \sum_{n=1}^\infty \frac{1}{n^x} .$$ But I thought that it did converge uniformly to some function $\xi(x):(1,\infty) \to \mathbb{R}$. Here's why; if you work on $(1+\delta,\infty)$ for $\delta >0$, then clearly we have $$\left\|\frac{1}{n^x}\right\|_\infty = \frac{1}{n^{1+\delta}}$$ and clearly $\sum \frac{1}{n^{1+\delta}}$ converges so by the Weierstrass M-test $\sum_{n=1}^\infty \frac{1}{n^x}$ converges uniformly on $(1+\delta,\infty)$ so letting $\delta \to 0$ gives $\sum_{n=1}^\infty \frac{1}{n^x}$ converging uniformly on $(1,\infty)$?

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No, you cannot do unions like that. Uniform convergence on each $(1+\delta,\infty)$ need not imply uniform convergence on $(1,\infty)$. But your work shows where to look for your counterexample: near $1$. – GEdgar Apr 11 '12 at 14:03
up vote 3 down vote accepted

Hint: Use an estimate that gives concrete information about the error when you cut off so that the first term left out involves $\frac{1}{m^x}$. This error is $\sum_m^\infty \frac{1}{n^x}$, which is greater than $$\int_{m}^\infty\frac{dt}{t^x}.$$ The above integral is equal to $$\frac{1}{x-1}\frac{1}{m^{x-1}}.\tag{$\ast$}$$ Now show that however large $m$ may be, there is an $x$ such that the expression in $(\ast)$ is well away from $0$. You might for example use $x=1+\frac{1}{m}$.

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