Alternating series test for complex series

• 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...)

Here, I thought it might be instructive to present an approach that uses a generalization of Dirichlet's Test and that has wider applicability. To that end we proceed.

Let $s\in \mathbb{C}$. The Dirichlet Eta function, $\eta(s)$, as represented by the series

$$\eta(s)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} \tag 1$$

is easily seen to converge for $\text{Re}(s)>1$.

If $s$ is purely real, then Dirichlet's test guarantees that the series representation converges for $\text{Re}(s)=s>0$. If $s$ is not purely real, then Dirichlet's test is inapplicable since the term $\frac{1}{n^s}$ is not real and monotonically decreasing.

It is not obvious, therefore, that the series in $(1)$ converges when $\text{Re}(s)>0$ for general complex values of $s$. There is a generalization of Dirichlet's test See Here to which we can appeal and show the convergence of $(1)$ whenever $\text{Re}(s)>0$.

The Generalized Dirichlet Test states that if $a_n$ and $b_n$ are, in general, complex sequences, then the sequence of their product, $\sum_{n=1}^\infty a_nb_n$, converges under the following three conditions:

1. There exists a number $M$, independent of $N$, such that the partial sums of $b_n$ satisfy

$$\left|\sum_{n=1}^N b_n\right|\le M$$

2. The terms $a_n$ tend to zero as $n\to \infty$

3. The sequence $a_n$ is of bounded variation with $\sum_{n=1}^\infty |a_{n+1}-a_n| \le L <\infty$

Let $a_n=\frac{1}{n^s}$ and $b_n=(-1)^{n-1}$, $\text{Re}(s)>0$. Clearly conditions $1$ and $2$ are satisfied. To show that $3$ is satisfied, we note that for fixed $s$ with $\text{Re}(s)>0$

$$\begin{align} \sum_{n=1}^\infty \left|\frac{1}{(n+1)^s}-\frac{1}{n^s}\right|& \le \int_1^\infty \left|\frac{d}{dt}\left(\frac{1}{t^s}\right)\right|\,dt\\\\ &=\int_1^\infty \left|\frac{-s}{t^{s+1}}\right|\,dt\\\\ &=|s|\int_1^\infty \frac{1}{t^{1+\text{Re}(s)}}\,dt\\\\ &=\frac{|s|}{\text{Re}(s)} \end{align}$$

where we recognized that $\sum_{n=1}^\infty \left|\frac{1}{(n+1)^s}-\frac{1}{n^s}\right|$ represented the sum of the lengths of secant lines of the parametric curve $\frac{1}{t^s}$, $t\in [1,\infty)$.

Therefore, by the Generalized Dirichlet Test, the series

$$\eta(s) =\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}$$

converges when $\text{Re}(s)>0$. And we are done!