Alternating series test for complex series

I want to show that we can continue Riemann's zeta function to Re$(s)>0$, $s\neq 1$ by the following formula

$$\begin{align} (1-2^{1-s})\zeta(s)&=\left(1-2\frac{1}{2^s}\right)\left(\frac1{1^s}+\frac1{2^s}+\ldots \right) \\ &=\frac1{1^s}+\frac1{2^s}+\ldots -2\left(\frac1{2^s}+\frac1{4^s}+\ldots \right)\\ &=\frac1{1^s}-\frac1{2^s}+\frac1{3^s}-\frac1{4^s}+\ldots \\ &=\sum_{n=1}^\infty (-1)^{n-1}\frac1{n^s}. \end{align}$$
In order to do that, I need to show that the series converges for Re$(s)>0$, except $s=\frac{2k\pi i}{\ln 2}+1$, $k\in \mathbb{Z}$, which are removable singularities. Any ideas on how to do that?

Answer

• Write the Taylor expansion of $(1+x)^a$ around $x= 0$, and

  $$n^{-s} - (n+1)^{-s} = n^{-s}(1- \left(\frac{n+1}{n}\right)^{-s}) = n^{-s}(1- \left(1+\frac{1}{n}\right)^{-s}) = \mathcal{O}(n^{-s-1})$$
  which is summable for $Re(s) > 0$, hence by grouping the terms by two :
  
  $$\eta(s) = (1-2^{1-s})\zeta(s) = \sum_{n=1}^\infty (2n-1)^{-s} - (2n)^{-s}$$ is absolutely convergent for any $Re(s) > 0$. ($\eta(s)$ is called the Dirichlet $\eta$ function )




• Other way, integrate by part (with $\delta(x)$ the Dirac delta distribution)

  $$\sum_{n=1}^\infty a_n n^{-s} = \int_{1-\epsilon}^\infty \left(\sum_{n=1}^\infty a_n \delta(x-n)\right) x^{-s} dx = s \int_1^\infty \left(\sum_{n \le x} a_n\right) x^{-s-1} dx$$ here $a_n = (-1)^{n+1}$ hence $\sum_{n \le x} a_n = 1$ or $0$ and $$\eta(s) = s \int_1^\infty \left(\sum_{n \le x} (-1)^{n+1}\right) x^{-s-1} dx$$ converges for $Re(s) > 0$ (it's called the Perron's formula )




• Last way, use that $n^{-s} \Gamma(s) = n^{-s} \int_0^\infty x^{s-1} e^{-x} dx = \int_0^\infty y^{s-1} e^{-ny} dy$ (change of variable $y = nx$, and $\Gamma(s)$ the Gamma function) hence :
  

  $$\Gamma(s) \sum_{n=1}^\infty a_n n^{-s} = \int_0^\infty y^{s-1} \left(\sum_{n=1}^\infty a_n e^{-ny} \right) dy$$
  (by using the absolute/dominated convergence theorem for exchanging $\sum$ and $\int$)
  
  
  

  here $\sum_{n=1}^\infty (-1)^{n+1} e^{-ny} = \frac{1}{e^y+1}$ hence
  

  $$\eta(s) \Gamma(s) = \int_0^\infty \frac{y^{s-1}}{e^y+1} dy$$
  which has no singularity for $Re(s) > 0$.




• Last last way :

  $$\eta(\epsilon) = \sum_{n=1}^\infty (-1)^{n+1} n^{-\epsilon}$$
  
  is a convergent alternated series for $\epsilon > 0$, hence by the abscissa of convergence theorem for Dirichlet series, we get that $\sum_{n=1}^\infty (-1)^{n+1} n^{-s}$ converges for any $Re(s) > \epsilon$, and since $\epsilon$ is arbitrary small, for any $Re(s) > 0$.

(if someone knows another way...)