Two trivial questions about zeta function I have two questions concerning the Riemann zeta function which are rather trivial so if anyone can give me the answers that would be nice, here is what I`m interested in:
1) In the equality $\zeta(s)=\sum_{n=1}^{\infty} \dfrac {1}{n^s}$ is the term $\dfrac {1}{n^s}$ by the definition equal to $e^{-s\log n}$, where $e$ is the natural logarithm base and $\log$ is the natural logarithm?
2) Because it is the case that it holds that $\zeta (s)={\dfrac {1}{1-2^{1-s}}}\sum_{n=1}^{\infty}\dfrac {(-1)^{n-1}}{n^s}$ is finding the zero of $\zeta (s)$ in the range $0<\Re(s)<1$ equivalent to finding the zero of $\eta (s)=\sum_{n=1}^{\infty}\dfrac {(-1)^{n-1}}{n^s}$ in the same range?
 A: Yes and yes.${}{}{}{}{}{}{}{}$
A: To offer a different perspective: while $\dfrac1{n^s}$ is generally equal by definition to $\dfrac1{e^{s\ln n}}$ and the function $e^t$ is a bit more canonical in some sense than the general power function, it doesn't really do any specific good to rework the definition.  In particular, it doesn't expand the domain of convergence of the series at all; both series are defined for $\Re(s)\gt 1$ and neither is convergent outside this domain.  Instead, to extend the domain of definition of the zeta function (and in particular to define it on the critical strip $0\lt \Re(s)\lt 1$), analytic continuation is used, for instance via the $\eta$-function you raise in point 2.
A: Eta function You mention is defined in $\Re s >0$ by the series You mention. Yes, zeta roots are also the roots of eta. Converse does not stand, because eta has some extra roots on line $\Re s =1$. 
A friendly note: series for eta converges conditionally only. This means, if one rearranges terms of the series, the result will change, and the sum will not necessarily be eta nor need to converge at all. Check http://en.wikipedia.org/wiki/Conditional_convergence and http://en.wikipedia.org/wiki/Riemann_series_theorem, for instance.
Yes, the usual real logarithm is fine.
Regards.
