Riemann Zeta approaches infinity as x approaches 1

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Homework Statement




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


Homework Equations



We cannot find the formula that [itex]\zeta[/itex] is given by, but we know
[itex]\sum_{n=1}^{\infty}\frac{1}{n^1}[/itex] 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 [itex]\zeta(x)[/itex] is continuous and can be differentiated.

The solution seem quite apparent: As [itex]x[/itex] approaches [itex]1[/itex], the series will approach the harmonic series, and thus it will diverge. Given any [itex]M>0[/itex], I can find a [itex]b>0[/itex] so that [itex]x \in (1,1+b) \Rightarrow \zeta(x) >M[/itex], 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 [itex]x[/itex] and then let [itex]n[/itex] approach infinity will be equivalent to letting n approach infinity and then let [itex]x[/itex] approach infinity? Or can I approach this in a different way?

Any help will be greatly appreciated!
 

Answers and Replies

  • #2
699
5
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##
$$
\infty = 1+\frac{k}{2}\leq \sum_{n=1}^{2^k}\frac{1}{n}
$$
 

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