Analytic continuation of Riemann zeta-function

Question

I showed that $$\zeta(s)=\frac1{1-2^{1-s}}\sum_{n=1}^\infty (-1)^{n-1}\frac1{n^s}.$$ Right hand side should be analytic when Re$(s)>0$ and $s\neq 1$, but there are problems at points $s=\frac{2k\pi i}{\ln 2}+1$, $k\in \mathbb{Z}$. How can we get rid of those singularities? Why does $s=1$ remain?

Answer 1

yes there is a problem at $s = 1+\frac{2 i k \pi}{\ln 2}, k \in \mathbb{Z}$. a common solution for solving it :

at first for $Re(s) > 1$, then by analytic continuation for $Re(s) > 0$ : $$\zeta(s) = \frac{1}{1-3^{1-s}}\sum_{n=0}^\infty (3n+1)^{-s} +(3n+2)^{-s}- 2 (3n+3)^{-s} = \frac{\eta_3(s)}{1-3^{1-s}}$$ where (exactly as $\eta(s) = \sum_{n=1}^\infty (-1)^{n+1} n^{-s}$) the function $\eta_3(s)$ is analytic on $Re(s) > 0$.

hence with this formulation, $\zeta(s)$ is analytic on $Re(s) > 0$ except at $s = 1+\frac{2 i k \pi}{\ln 3}, k \in \mathbb{Z}$, and since $2$ and $3$ are relatively primes, the intersection between $\{ 1+\frac{2 i k \pi}{\ln 2} \ \mid \ k \in \mathbb{Z}\} $ and $\{1+\frac{2 i k \pi}{\ln 3} \ \mid \ k \in \mathbb{Z}\}$ reduces to $\{1\}$, and hence $\zeta(s)$ is analytic on $Re(s) > 0, s \ne 1$.

this is also the way for proving that $\eta(1+\frac{2 i k \pi}{\ln 2}) = 0$ for every $ k \in \mathbb{Z}\setminus \{0\}$.

(this is explained on https://fr.wikipedia.org/wiki/Fonction_zêta_de_Riemann, or you can look at https://en.wikipedia.org/wiki/Dirichlet_eta_function)

from the fact that $1-2^{1-s}$ has a zero of order $1$ at $s=1$, we also get that $(s-1)\zeta(s)$ is analytic at $s=1$, and since $$\lim_{\sigma \to 1^+} \zeta(\sigma) = \lim_{\sigma \to 1^+} \sum_n n^{-\sigma} = + \infty$$

we find that $\zeta(s)$ is not analytic at $s=1$, hence it has a pole of order $1$ there.

Answer 2

Consider the series $$F(s)=\frac{1}{1^s}+\frac{1}{2^s}-\frac{2}{3^s}+\frac{1}{4^s}+\frac{1}{5^s}-\frac{2}{6^s}+\cdots$$ First, convergence: $$ \begin{eqnarray} F(s)&\leq&\frac{1}{1^s}+\frac{1}{1^s}-\frac{2}{3^s}+\frac{1}{4^s}+\frac{1}{4^s}-\frac{2}{6^s}+\cdots \\ &=&\frac{2}{1^s}-\frac{2}{3^s}+\frac{2}{4^s}-\frac{2}{6^s}+\frac{2}{7^s}\cdots \end{eqnarray} $$ This converges for $\mathfrak{R}(s)\geq0$ for the same reasons the series in your formula does.

One quickly checks that $\zeta(s)-F(s)=\frac{3}{3^s}\zeta(s)$, and so $$\zeta(s)=\frac{1}{1-3^{1-s}}F(s)$$ Thus we have an analytic continuation of $\zeta$ with poles at $s=1+k\frac{2\pi i}{\ln(3)}$. The only common pole shared with your formula is thus at $s=1$. Together, the two formulas show that $\zeta$ is analytically continuable to $s>0$, except at $s=1$

At $s=1$, no such trick will work. To see that $\zeta$ has an actual pole there, it suffices to show that the residue of $\zeta$ at $s=1$ is non-zero. Using the formula you quote: $$ \begin{eqnarray} \mathrm{Res}(\zeta,1) &=&\lim_{s\rightarrow 1}(s-1)\zeta(s) \\ &=&\lim_{s\rightarrow 1}\frac{(s-1)}{1-2^{1-s}}\sum_{n=1}^\infty(-1)^n\frac{1}{n^s} \\ &=&\lim_{s\rightarrow1}\frac{(s-1)}{1-2^{1-s}}\ln(2) & \textrm{(Taylor series of }\ln(1+s))\\ &=&\lim_{s\rightarrow1}\frac{1}{(\ln{2})2^{1-s}}\ln(2) & \textrm{(L`Hospital's rule)} \\ &=& 1 \end{eqnarray} $$