How to prove the Riemann Zeta fuction tends to infinity when $x$ tends to $1$

Question

The Riemann Zeta Function is convergent over the interval $(1,\infty)$, and $\sum_{n=1}^{\infty}\frac{1}{n^x}$ tends to infinity when $x\rightarrow 1^ {+} $, it seems one can feel it is right because the when $x=1$ the function is infinite and it is monotonely decreasing over the interval $(1,\infty)$.

But since $\sum_{n=1}^{\infty}\frac{1}{n^x}$ is not uniformly convergent at the interval $(1,1+\delta)$, so I can't simply exchange the limit order as follows: $$\lim_{x\rightarrow1}\sum_{n=1}^{\infty}\frac{1}{n^x}=\sum_{n=1}^{\infty}\frac{1}{n}$$, so how can I compute this: $$\lim_{x\rightarrow1}\sum_{n=1}^{\infty}\frac{1}{n^x}$$

Answer 1

We show that for every $M \in \Bbb R_{>0}$, there exists $\delta > 0$ such that $\zeta(x) > M$ for all $x \in (1, 1+\delta)$.

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Assume that such an $M$ is given. As $\sum 1/n$ diverges, there exists $N \in \Bbb N$ such that $$\sum_{n=1}^N\dfrac{1}{n} > 2M.$$

Also, note that for each $n \in \{1, \ldots, N\}$, we have a $\delta_n > 0$ such that $$\dfrac{1}{2n} < \dfrac{1}{n^x} , \qquad (*)$$ for all $x \in (1, 1+\delta_n).$
(This is because $n^x \to n$ as $x\to1^+$.)

By choosing $\delta = \min\{\delta_n \mid n \in \{1, \ldots, N\}\}$, we see that $(*)$ holds for all $x \in (1, 1 + \delta)$ and for all $n\in\{1,\ldots,N\}$.

Thus, we have \begin{align} \sum_{n=1}^N\dfrac{1}{n^x} &> \sum_{n=1}^N\dfrac{1}{2n}\\ &> \dfrac{1}{2}(2M)\\ &= M, \end{align} for all $x \in (1, 1 + \delta)$.

This gives us that \begin{align} \zeta(x) &= \sum_{n=1}^\infty \dfrac{1}{n^x}\\ &>\sum_{n=1}^N \dfrac{1}{n^x}\\ &> M, \end{align} for all $x \in (1, 1 + \delta)$, as desired.

Answer 2

You can bound the sum by an integral that goes to infinity as $x\to1$.

Answer 3

For each $N$ we have $ \sum\limits_{n=1}^{\infty} \frac 1 {n^{x}} \geq \sum\limits_{n=1}^{N} \frac 1 {n^{x}}$ and hence $\lim \inf_{x \to 1} \sum\limits_{n=1}^{\infty} \frac 1 {n^{x}} \geq \lim \inf_{x \to 1} \sum\limits_{n=1}^{N} \frac 1 {n^{x}} =\sum\limits_{n=1}^{N} \frac 1 {n} $. Can you finish?