# Riemann Zeta approaches infinity as x approaches 1

1. May 13, 2012

### pndc

1. The problem statement, all variables and given/known data

We know $\sum_{n=1}^{\infty}\frac{1}{n^x}$ is uniformly convergent on the interval $x\in(1,\infty)$ and that its sum is called $\zeta(x)$. Proof that $\zeta(x) \rightarrow \infty$ as $x \rightarrow 1^+$.

2. Relevant equations

We cannot find the formula that $\zeta$ is given by, but we know
$\sum_{n=1}^{\infty}\frac{1}{n^1}$ diverges.

3. The attempt at a solution
I've already shown that the functional series is uniformly convergent on the above interval and that the derivative also converges uniformly to a limit function, thus $\zeta(x)$ is continuous and can be differentiated.

The solution seem quite apparent: As $x$ approaches $1$, the series will approach the harmonic series, and thus it will diverge. Given any $M>0$, I can find a $b>0$ so that $x \in (1,1+b) \Rightarrow \zeta(x) >M$, as we know that we will approach the harmonic series.

However, I'm not quite sure how I can go about formulating a formal proof. For functional series, I suppose I cannot just plug in the limit and say its a harmonic series (especially as it's not defined there).

Under which conditions can I be completely sure that plugging in the relevant $x$ and then let $n$ approach infinity will be equivalent to letting n approach infinity and then let $x$ approach infinity? Or can I approach this in a different way?

Any help will be greatly appreciated!

2. May 17, 2012

### Dustinsfl

Why not use the fact that
$$\sum_{n=1}^{\infty}\frac{1}{n}$$
is a harmonic series.

Then
$$1+\frac{k}{2}\leq\sum_{n=1}^{2^k}\frac{1}{n}$$
as $k\to\infty$
$$\infty = 1+\frac{k}{2}\leq \sum_{n=1}^{2^k}\frac{1}{n}$$