I know values of zeta for s= 2,3,4,... but what's the value of zeta as an example s= 2+14i
Does Zeta converge for Complex numbers Re(s)>1 where imaginary part is not zero?
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0Yes 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 $s=3$.) – 2012-10-17
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0Probably Mr Chuy should spend some more time working on analysis ... it will surely help him if he wants to work on the zeta function. – 2012-11-08
4 Answers
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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0Thanks everbody,all the answers were helpful. – 2012-10-17
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 $\sigma > 1$.
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,
The imaginary part of $s$ is irrelevant for the convergence of the series. Just notice that
$$ |k^{-s}| = |e^{-s\ln(k)}| = |e^{-(u+iv)\ln(k)}| = e^{-u\ln(k)} = k^{-u}\,. $$