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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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Side note: The root test doesn't help here ($\limsup_{n\to\infty}\sqrt[n]{|a_n|} = 1$). In fact, this is mentioned in the Wikipedia article you linked to. –  Ayman Hourieh Sep 21 '12 at 16:02
$s_\epsilon \rightarrow \frac1\epsilon$ –  karakfa Sep 21 '12 at 16:06
@karakfa: What do you mean? –  Did Sep 21 '12 at 16:11
@karakfa: Does your answer mean anything else than "$s_\varepsilon$ diverges for $\varepsilon \rightarrow 0$"? –  moose Sep 21 '12 at 16:12
Well, but many systems will have numerical algorithms for the zeta function ... –  kjetil b halvorsen Sep 21 '12 at 17:17

2 Answers 2

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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Well, the Riemann zeta function is exactly what you are asking about. It doesn't answer your question. You can now rephrase your question, if you wanted, to ask what is the value of $\zeta(1 + \epsilon)$ for $\epsilon > 0$. The answer is that there are some values of $\epsilon$ where there is a known answer, for instance $\zeta(2n)$ is known for all positive integers $n$. But, $\zeta(2n+1)$ is not known in any nice way for any $n \geq 1$. You simply have a name for what you are asking, so it's easier to look things up, but you still don't know the answer for almost all $\epsilon$. –  Graphth Sep 21 '12 at 15:58
Knowing that it is not easy to find an answer is still much more than I knew before I've asked the question. And knowing the name is very important useful if you want to find an answer by yourself. –  moose Sep 21 '12 at 16:09
I know, I'm just saying it doesn't really answer your question as you claim in your answer, "So the answer is: ...". But, yet, knowing the name will help you get more answers more easily. –  Graphth Sep 21 '12 at 16:12
Ok, you're right. I've edited this part. –  moose Sep 21 '12 at 16:18

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