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Does $\sum_{k=1}^\infty \frac{1}{k^x} $ converge uniformly on $\left(\sqrt{2}, \infty\right)$?

Can somebody tell me if I am doing this correctly? If not, what am I doing wrong? It is clear immediately that this series converges. I dont think the Weierstrass M-Test (WMT) will work since it is my $x$ that varies. To show convergence is uniform, it suffices to show that

$\text{sup}_{x\in \left(\sqrt{2},\infty\right)}\left|f_n-f\right| \to 0 \text{ as } n\to\infty.$ (Rudin 7.9)

$\left|f_n - f\right| = \left|\sum_{k=N+1}^\infty \frac{1}{k^x}\right|$. From here I can consider sup$\left|\sum_{k=N+1}^\infty \frac{1}{k^x}\right|=1.$ Since this doesn't converge, we can see that the sum doesn't converge uniformly.

I'm unsure if this is valid, can somebody help?

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  • $\begingroup$ M-test and $p$-series solves the problem. $\endgroup$ – Mhenni Benghorbal Dec 9 '13 at 4:52
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Since $x>\sqrt2$, $k^x>k^{\sqrt2}$ for all $k\in\mathbb N$. Then $$ \left|\sum_{k=N+1}^\infty\frac1{k^x}\right|=\sum_{k=N+1}^\infty\frac1{k^x} \leq\sum_{k=N+1}^\infty\frac1{k^{\sqrt2}}. $$ So the convergence of $\sum_kk^{-\sqrt2}$ implies that the convergence of your series is indeed uniform.

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  • $\begingroup$ So when I use the M-Test I am not require to choose an $x\in \left(\sqrt{2}, \infty\right)$? $\endgroup$ – Kyle H. Dec 9 '13 at 4:44
  • $\begingroup$ The point of the M-test is to compare with a series that does not depend on $x$, and the bound should work for all $x$. $\endgroup$ – Martin Argerami Dec 9 '13 at 6:05

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