Is there any value which we can substitute for $s$ in $\zeta (s)$ such that $$\sum_{n=1}^{\infty }n^{-s}\in \mathbb{Z}$$

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    $\begingroup$ Note that, $\lim_{s\to \infty}\zeta(s)=1 $. $\endgroup$ – Mhenni Benghorbal Aug 14 '13 at 1:44
  • $\begingroup$ I thought $\zeta (s)$ increases as $\lim_{n \to \infty }$ because $\zeta (2) \approx 1.6$ and $\zeta (4) \approx 10$. Interesting :D $\endgroup$ – please delete me Aug 14 '13 at 1:47
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    $\begingroup$ @user2357 That's because it's not injective (aka it changes "direction"). $\endgroup$ – Ataraxia Aug 14 '13 at 1:51
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    $\begingroup$ @Ataraxia On the interval $(1,\infty)$ where the sum representation is convergent, $\zeta$ should be decreasing because each term in the sum is. And $\zeta(4)\approx 1.08$. See specific values. $\endgroup$ – Jeppe Stig Nielsen Aug 14 '13 at 6:11
  • $\begingroup$ Oh, yes, i had a 9 instead of the 90 lol. Thank you! $\endgroup$ – please delete me Aug 14 '13 at 6:17

$\zeta(s)$ is a continuous function with $\lim_{s \rightarrow \infty} \zeta(s) = 1$ and $\lim_{s \rightarrow 1} \zeta(s) = \infty$ (on the real line), so it takes every positive integer at some $s > 1$. Just the intermediate value theorem.

  • $\begingroup$ How do you know it's continuous? $\endgroup$ – Ataraxia Aug 14 '13 at 1:47
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    $\begingroup$ Many ways.. it's analytic for one. More elementarily you can look at $\zeta(s + \delta) - \zeta(s)$. Given $\epsilon$ and $N$, if $\delta$ is small enough the difference of the first $N$ terms can be made less than ${\epsilon \over 2}$, and also the sum of the absolute values of the remaining terms will be less than ${\epsilon \over 2}$ if $N$ were chosen large enough $\endgroup$ – Zarrax Aug 14 '13 at 1:50
  • $\begingroup$ Is it not possible for $\zeta(s)$ to evaluate to an integers for $s<\infty $. Is it possible to prove that it doesn't? $\endgroup$ – please delete me Aug 14 '13 at 1:55
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    $\begingroup$ @user2357 Do you know the intermediate value theorem? $\endgroup$ – Ataraxia Aug 14 '13 at 1:57
  • $\begingroup$ Every positive integer, except $\zeta=1$, unless you allow $\zeta(+\infty)$. $\endgroup$ – Jeppe Stig Nielsen Aug 14 '13 at 6:15

When we analytically continue $\zeta$ to values of $s$ where the series doesn't converge, we get the famous Riemann zeta function, for which $\zeta(-2)=0$. That might be the easiest integer value to write down!


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