Originally Posted by

**robinlrandall** Thanks SlipEternal,

I think I have to understand better how Hardy move zeta(s) into the Complex Plane. Could $\displaystyle \sum_{k=1}^\infty \dfrac{1}{k^{1/2 +it}} $ converge to 0?

Possibly, if we are in the Complex Plane. I need to do a bit more research. But thanks again. Oh, on $\displaystyle \sum_{k=1}^\infty \dfrac{1}{k^{1/2}}$ I hadn't really thought of it that way, but perhaps I concluded similarly. Thanks again.

Better understanding of below should help:

The Riemann hypothesis discusses zeros outside the region of convergence of this series, so it must be analytically continued to all complex s. This can be done by expressing it in terms of the Dirichlet eta function as follows. If the real part of s is greater than one, then the zeta function satisfies

$\displaystyle \left(1-\frac{2}{2^s}\right)\zeta(s) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s} = \frac{1}{1^s} - \frac{1}{2^s} + \frac{1}{3^s} - \cdots.$

However, the series on the right converges not just when s is greater than one, but more generally whenever s has positive real part. Thus, this alternative series extends the zeta function from $\displaystyle \text{Re}(s) > 1$ to the larger domain $\displaystyle \text{Re}(s) > 0$, excluding the zeros $\displaystyle s = 1 + 2\pi in/\ln(2)$ of $\displaystyle 1-2/2^s$ (see Dirichlet eta function). The zeta function can be extended to these values, as well, by taking limits, giving a finite value for all values of s with positive real part except for a simple pole at $\displaystyle s = 1$.

In the strip $\displaystyle 0 < \text{Re}(s) < 1$ the zeta function satisfies the functional equation

$\displaystyle \zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s)$.

One may then define ζ(s) for all remaining nonzero complex numbers s by assuming that this equation holds outside the strip as well, and letting ζ(s) equal the right-hand side of the equation whenever s has non-positive real part. If s is a negative even integer then ζ(s) = 0 because the factor sin(πs/2) vanishes; these are the trivial zeros of the zeta function. (If s is a positive even integer this argument does not apply because the zeros of sin are cancelled by the poles of the gamma function as it takes negative integer arguments.) The value ζ(0) = −1/2 is not determined by the functional equation, but is the limiting value of ζ(s) as s approaches zero. The functional equation also implies that the zeta function has no zeros with negative real part other than the trivial zeros, so all non-trivial zeros lie in the critical strip where s has real part between 0 and 1.

Robin