Does Zeta converge for Complex numbers Re(s)>1 where imaginary part is not zero? I know values of zeta for s= 2,3,4,... but what's the value of zeta as an example s= 2+14i
 A: 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 $\zeta(3)$. 
A: 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 $\sigma > 1$.
A: 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,
A: The imaginary part of $s$ is irrelevant for the convergence of the series. Just notice that
$$ |k^{-s}| = |e^{-s\ln(k)}| =  |e^{-(u+iv)\ln(k)}| = e^{-u\ln(k)} = k^{-u}\,. $$ 
