# Some clarifications on analytic continuation of Riemann's Zeta function on $\frac 1 2$

Here's my problem: Riemann's Zeta function converge iff $x>1$ so if I want to have a finite value for $\zeta(\frac 1 2)$ I need to use it's analytic continuation but Riemann's hypothesis states that every non-trivial zero of this function lies on the critical line of equation $x=\frac 1 2 +in$ so if I set $n=0$ I get that $\zeta(\frac 1 2)$ should be $0$ and I thought I would have to use a similar way to the one that gets me $0$ for every even negative integer but Wikipedia says that:

$\zeta(\frac 1 2)=\sum_{n=1}^\infty \frac 1 {\sqrt n}=1+\frac 1 {\sqrt2}+\frac 1 {\sqrt3}+\dots\approx-1,4603545$

1)how can I get to this result ?

2)Isn't there any closed form (like the one involving Bernoulli numbers for every non-positive integer) or only approximated ones ?

3)Can I apply this method to other rational numbers <1 ? (es. $\frac 1 4$ or $-\frac 1 2$)

• There are several important inaccuracies here: the zeta function converges for compelx with real part greater than one only if we take the basic, most well know definition of it, namely: $$\zeta(s)=\sum_{n=1}^\infty\frac1{n^s}\;\;,\;\;\;\text{Re}\,s>1$$ . If we want to deal for example with $\;s\in\Bbb C\;,\;\;0\le\text{Re}\,s<1\;$ we have to analitically continue the above, say by means of Dirichlet's Alternating Eta Function. Now, R-H says all zeros of the zeta fun. are on the strip $\;\text{Re}\,s=\dfrac12\;$ , but this does not mean any such number must be a zero. – Timbuc Mar 31 '15 at 17:18
• For the first question, see math.stackexchange.com/questions/947158/… and math.stackexchange.com/questions/1082139/… – Chris Culter Mar 31 '15 at 17:18
• Thanks @Timbuc now I got less confusion in my mind – Renato Faraone Mar 31 '15 at 17:21

## 2 Answers

Yes, at least some zeros of $\zeta(s)$ lie on the line $\Re(s)=1/2$, but they occur at irregular intervals (this is one of the most important features of the zeta function in the critical strip), and in particular $\zeta(1/2) \neq 0$ (the first one occurs at $s=\frac{1}{2} \pm 14.\text{something}$).

Your series diverges (as does the normal Dirichlet series for any real $s$ with $\Re(s) \leq 1$), but you can use the identity $$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^s} = (1-2^{1-s})\zeta(s),$$ where the series on the left converges for $\Re(s)>0$ using the alternating series test. (See the links @ChrisCulter posted in the comments for a proof.)

There is no known closed form, but another option is to approximate the integral $$\frac{1}{(1-\sqrt{2})\sqrt{\pi}}\int_0^{\infty} \frac{x^{-1/2}}{e^x+1} \, dx,$$ which also evaluates to $\zeta(1/2)$.

To the formula: $\zeta(s)=\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^s}$

is written in small print: $1<\Re(s)$

That's a classical trap. Here my solution:

According to the russian Wikipedia (this have something in common with the Euler–Maclaurin formula) :

$\zeta(s)=\displaystyle\lim_{N \to \infty}(\displaystyle\sum_{n=1}^{N} \frac{1}{n^s} -\frac{N^{1-s}}{1-s})$ ;for $\Re(s)>0$ and $\Re(s)\ne 1$.

By putting s=1/2 , we get:

$\zeta(1/2)=\displaystyle\lim_{N \to \infty}(\displaystyle\sum_{n=1}^{N} \frac{1}{\sqrt{n}} -2\sqrt{N})$

This series converges very slowly but converge.