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I need to prove the following:

$$\lim_{s \to \infty} \sum_{n = 1}^\infty \frac 1 {n^s} = 1$$

This is my attempt:

\begin{align} \lim_{s \to \infty} \sum_{n = 1}^\infty \frac 1 {n^s} & = \lim_{s \to \infty} \left( 1 + \sum_{n = 2}^\infty \frac 1 {n^s} \right) \\ & = 1 + \sum_{n = 2}^\infty \lim_{s \to \infty} \frac 1 {n^s} \\ & = 1 + \sum_{n = 2}^\infty 0 \\ & = 1 \end{align}

I need help justifying the interchange of the sum and the limit though. Is it enough to show the limit exists and the sum is convergent? I read that I need to prove uniform convergence, how can I do this?

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    $\begingroup$ Hint : Consider $\int \frac{1}{x^s}\ dx$ $\endgroup$ – HK Lee May 8 '14 at 6:10
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Let $s\gt 1$, and let our sum be $A(s)$. A picture shows that $$1\lt A(s)\lt 1+\int_1^\infty \frac{1}{x^s}\,dx.$$ But $\int_1^\infty \frac{1}{x^s}\,dx$ can be explicitly computed, and approaches $0$ as $s\to\infty$.

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For $s\ge 2$ we have $$\frac1{n^s}\le \frac1{n^2}$$ and the series $$\sum_{n=1}^\infty \frac1{n^2}$$ is convergent and doesn't depend on $s$ so the given series is uniformly convergent and you can interchange limit and sum.

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