Understanding Riemann Zeta functions for s=1/3

[joao_pimentel]

Hi guys

I'm trying to understand Riemann Zeta functions particularly for s=1/3

I know $$\zeta(s)=\sum_{n=1}^\infty\frac{1}{n^s}$$

and converges for Re(s)>1

Ok, but what about for s=1/3, then

$$\zeta(1/3)=\sum_{n=1}^\infty\frac{1}{n^{1/3}}=<br /> \sum_{n=1}^\infty\frac{1}{\sqrt[3]{n}}$$

Theoretically it should not converge, but when I put $$\zeta(1/3)$$ on Wolfram I get approximately -0,9

May you kindly explain me this result, please?

________________________

Why? Because I'm trying to solve this problem:
$$\left[\frac{1}{\sqrt[3]{4}}+\frac{1}{\sqrt[3]{5}}+\frac{1}{\sqrt[3]{6}}+...+\frac{1}{\sqrt[3]{1000000}}\right]$$

where $$\left[x\right] =$$ is the Greatest Integer function

And I thought Riemann Zeta functions might be the solutions...

Any tips?

Thank a lot in advance guys for your support...

 

[Char. Limit]

Wolfram-Alpha gives that result mainly because the Riemann Zeta Function formula you're using, i.e.

$$\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}$$

Only applies when Re(n)>1. Since the number you're using doesn't fit that application, we have to use a different formula, usually this one:

$$\zeta(s) = \frac{1}{\Gamma(s)} \int_0^\infty \frac{x^{s-1}}{e^x - 1} dx$$

That formula should give you a solution to Zeta(1/3). But I don't think it'll apply in this problem.

 

[joao_pimentel]

ok, thank you so very much for your attention...

I'll continue thinking on the subject...

thanks