Problem with understanding the analytic continuation of zeta function. I know two things:


*

*$\sum_{n=1}^{\infty}\frac{1}{n^{s}}$ converges if and only if $Re s > 1$

*analytic continuation is expanding the domain of entire function.


I would like to find analytic continuation of Riemann zeta function.
But there is a problem, as we approach $Re s=1$ from right side the value of series goes to infinity. Then how it is possible to find a smooth crossing to the $Re s \le 1$ part, so that we would have formula calculating zeta function for all complex numbers?(expect case when $s=1$)
 A: Analytic continuation doesn't work by somehow making $\sum_{n=1}^\infty\frac1{n^s}$ sum for additional values of $s$. But $\zeta(s)$ is an analytic function. So for any $z_0$ with $\Re(z_0) > 1$, we can also describe it by $$\zeta(z) = \sum_{n=0}^\infty \zeta^{(n)}(z_0)(z - z_0)^n$$
Now, the radius of convergence of this series is the distance from $z_0$ to the nearest non-analytic point for $\zeta$, which turns out to be $z = 1$.
So for example, if we take $z_0 = 4 + 4i$, then the distance from $z_0$ to $1$ is $\sqrt{(4-1)^2 + 4^2} = 5$. This gives us an expression for $\zeta$ that defines it everywhere within a distance of $5$ from $4+4i$. This includes $z_1 = 1+4i$, well within that boundary. so even though $\Re(z_1) = 1$, we still know that $\zeta$ is defined and analytic there. The nearest singularity to $z_1$ is still $1$, so the Taylor series about $z_1$ converges for a radius of $4$. This includes the point $z_2 = -2+4i$, so we can take another Taylor series about it...
That is analytic continuation. You start with the region where the function was originally defined, and take Taylor series about points in it. Depending on the function, one generally finds that the Taylor series about some points in the known domain actually converge on disks that extend beyond that domain, and so we can extend the domain to cover all of those disks. And it is likely that some of the new points in this extended domain likewise have radii of convergence that extends out beyond that new domain, allowing the same thing to happen again. For the Riemann zeta function, this works everywhere except at $z = 1$, the only pole of the function.
