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I know values of zeta for s= 2,3,4,... but what's the value of zeta as an example s= 2+14i

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Yes the series $\sum_{n=1}^\infty n^{-s}$ converges whenever $\mathrm{Re}(s)>1$, regardless of the imaginary part of $s$. A standard comparison test should suffice. Also, the body of your question seems to be asking for explicit values of the zeta function at nonreal arguments. Do you want numerical values (which e.g. WolframAlpha can spit out easily), or closed-form symbolic expressions? (If the latter, I have a hard time believing you know the value of $\zeta$ for $s=3$.) –  anon Oct 17 '12 at 0:43
    
Probably Mr Chuy should spend some more time working on analysis ... it will surely help him if he wants to work on the zeta function. –  GEdgar Nov 8 '12 at 15:55

4 Answers 4

What do you mean when you say you "know" the value of $\zeta(3)$? There is a sense in which nobody knows this value --- no one knows a finite expression for it in terms of square roots, cube roots, exponentials, logarithms, trig functions, $\pi$, and so on. And there is a sense in which everybody knows it: it the number to which the infinite series $\sum_1^{\infty}n^{-3}$ converges.

Well, the value of $\zeta(2+14i)$ is just the number to which $\sum_1^{\infty}n^{-2-14i}$ converges --- we know it just as well as, and in the same way as, we know $\zeta(3)$.

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Thanks everbody,all the answers were helpful. –  Chuy Oct 17 '12 at 0:51

The $\zeta$-functions is analytic on $\mathbb{C} \backslash\{1\}$. Hence, it converges for all $z \in \mathbb{C} \backslash\{1\}$.

Your question should probably reworded as

Does $\displaystyle \sum_{n=1}^{\infty} \dfrac1{n^s}$ converge for $\text{Real}(s) > 1$?

The statement is true and the proof is rather trivial since if $s = \sigma + it$, where $\sigma > 1$, we get $$\left \vert \displaystyle \sum_{n=1}^{N} \dfrac1{n^s} \right \vert \leq \displaystyle \sum_{n=1}^{N} \left \vert \dfrac1{n^s} \right \vert = \displaystyle \sum_{n=1}^{N} \left \vert \dfrac1{n^{\sigma + it}} \right \vert = \displaystyle \sum_{n=1}^{N} \dfrac1{n^{\sigma}} \left \vert \dfrac1{n^{it}} \right \vert = \displaystyle \sum_{n=1}^{N} \dfrac1{n^{\sigma}}$$ Now take the limit as $N \to \infty$ and recall that $\displaystyle \sum_{n=1}^{N} \dfrac1{n^{\sigma}}$ converges for $\sigma > 1$.

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Your question title asks if $\zeta(s)$'s series converges when $\mathrm{Re}(s) > 1$ and $\mathrm{Im}(s) \neq 0$ but your question asks differently, that's an issue.

About existence, you should know this : $$ \left| \sum_{n = 1}^{N} \frac 1{n^s} \right| \le \sum_{n = 1}^{N} \frac 1{n^{\mathrm{Re}(s)}} $$ because $$ \left| \frac 1{n^s} \right| = \left| \frac 1{e^{s \log n}} \right| = \left| \frac 1{e^{\mathrm{Re}(s) \log n}} \frac 1{e^{i\mathrm{Im}(s) \log n}} \right| = \left| \frac 1{e^{\mathrm{Re}(s) \log n}} \right| = \left| \frac 1{n^{\mathrm{Re}(s)}} \right|. $$ Therefore the series for $\zeta(s)$ converges when $\zeta(\mathrm{Re}(s))$ converges, hence the region $\mathrm{Re}(s) > 1$ (after some working out of the details in the real $s$ case).

Hope that helps,

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The imaginary part of $s$ is irrelevant for the convergence of the series. Just notice that

$$ |k^{-s}| = |e^{-s\ln(k)}| = |e^{-(u+iv)\ln(k)}| = e^{-u\ln(k)} = k^{-u}\,. $$

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