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I know that

$\displaystyle \sum_{n=1}^\infty x^n = \frac{1}{1-x}$ (Geometric series)

and that the harmonic series is divergent:

$\displaystyle \sum_{n=1}^\infty \frac{1}{n} \rightarrow \infty$

And I see quite often series of this form:

$s_\varepsilon =\displaystyle \sum_{n=1}^\infty \frac{1}{n^{1+\varepsilon}}$ with $\varepsilon > 0$

I know that $s_{\varepsilon > 0}$ converges due to the root test. But what is the value of those series?

So here is my question:

Let $\varepsilon > 0$. What is the value of $s_\varepsilon := \displaystyle \sum_{n=1}^\infty \frac{1}{n^{1+\varepsilon}}$?

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    The value of that sum is the Riemann zeta function evaluated at $1+\varepsilon$.2012-09-21

2 Answers 2

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Your answer is of course correct that you are looking at the $\zeta$ function. You don't mention $\epsilon$ being small, but you may also be interested in an expansion in small $\epsilon$, $\sum_{n=1}^\infty\frac{1}{n^{1+\epsilon}}=\frac{1}{\epsilon}+\gamma_0-\gamma_1\epsilon+O\left(\epsilon^2\right)$ with the $\gamma_i$ called Stieltjes constants.

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As I was looking for an answer, I've noticed:

According to Wolfram|Alpha:

  • $s_1 = \frac{\pi^2}{6}$ (Source) $= \zeta(2)$ (Source)
  • $s_\frac{5}{4} = \zeta(\frac{5}{4})$ (Source)
  • $s_{1+0.25} = 4.59511$ (Source)

This series has a name; it is the Riemann zeta function, which is exactly defined like this:

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

An expicit form for the value of $\zeta(s)$ seems to be unknowen.

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    Ok, you're right. I've edited this part.2012-09-21