# Is there any value of zeta that is an integer?

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

• Note that, $\lim_{s\to \infty}\zeta(s)=1$. – Mhenni Benghorbal Aug 14 '13 at 1:44
• I thought $\zeta (s)$ increases as $\lim_{n \to \infty }$ because $\zeta (2) \approx 1.6$ and $\zeta (4) \approx 10$. Interesting :D – please delete me Aug 14 '13 at 1:47
• @user2357 That's because it's not injective (aka it changes "direction"). – Ataraxia Aug 14 '13 at 1:51
• @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. – Jeppe Stig Nielsen Aug 14 '13 at 6:11
• Oh, yes, i had a 9 instead of the 90 lol. Thank you! – 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.
• 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 – Zarrax Aug 14 '13 at 1:50
• Is it not possible for $\zeta(s)$ to evaluate to an integers for $s<\infty$. Is it possible to prove that it doesn't? – please delete me Aug 14 '13 at 1:55
• Every positive integer, except $\zeta=1$, unless you allow $\zeta(+\infty)$. – 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!