How can I prove that $\lim_{x \rightarrow 1}{\sum_{n=1}^{\infty}{\frac{1}{n^x}}} = \infty$? My idea is to show that we can exchange the positions of limit and sum, obtaining the harmonic sum, that we know that diverges, but, are we able to do such exchange? If so, which results does allow us to do that?
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1For example, you can appeal to the monotone convergence theorem. – Daniel Fischer Jun 22 '15 at 12:59
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@DanielFischer Yes, we all love sledgehammers. I suggest splitting the sum into blocks between $(2^k)^{-x}$ and $(2^{k+1})^{-x}$, and showing that these are all larger than $1/2$. It's much more basic (basically it's an implementation of the condensation test). OTOH, surely this has an answer on here already? – Chappers Jun 22 '15 at 13:12
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This is a candidate for a duplicate. This one is probably overkill, but nevertheless interesting. – Daniel Fischer Jun 22 '15 at 13:28
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3Possible duplicate of Limit $(t-1)\zeta(t)$ as $t\rightarrow 1^+$ – Nosrati Dec 18 '18 at 16:09
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You may compare the sum with an integral : Since $t \mapsto \frac{1}{t^s}$ is decreasing we have
$$\frac{1}{n^s} \ge \int_{n}^{n+1} \frac{dt}{t^s}$$
From which we get
$$\sum_{n = 1}^{\infty} \frac{1}{n^s} \ge \int_{1}^{\infty} \frac{dt}{t^s} = \frac{1}{1-s} \underset{s \to 1}{\longrightarrow} + \infty$$
Actually by being more careful (using $\int_{n}^{n+1} \frac{dt}{t^s} \ge \frac{1}{(n+1)^s}$ on the other side) we'd get an equivalent of $\sum_{n = 1}^{\infty} \frac{1}{n^s}$ at $1$.
Joel Cohen
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