Riemann Zeta approaches infinity as x approaches 1

[pndc]

Homework Statement

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^+.

Homework Equations

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

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!

 

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