Show that $\sum_{n=1}^\infty\frac{(-1)^n}{n^s}$ converges conditionally for $Re(s)>0$

Question

I thought about doing the ratio test: $$\lim_{n\to\infty}\big|\frac{(-1)^{n+1}}{(n+1)^s}*\frac{n^s}{(-1)^n}\big|=\lim_{n\to\infty}\frac{n^s}{(n+1)^s}$$ Now, since $s$ is complex I am not sure on how to solve this problem. Thanks for any help

Answer 1

The usual argument involves "partial summation". Let $a_n=\sum_{k=1}^n(-1)^k$. So $a_n$ is either $-1$ or $0$, but $(-1)^n=a_n-a_{n-1}$. Then $$\sum_{n=1}^N\frac{(-1)^n}{n^s}=\sum_{n=1}^N\frac{a_n-a_{n-1}}{n^s}= \sum_{n=1}^{N-1}a_n\left(\frac1{n^s}-\frac1{(n+1)^s}\right)+\frac{a_N}{N^s}.$$ Let $\sigma=\text{Re}(s)>0$. Then $a_N/N^s\to0$ as $N\to\infty$. Moreover $$\left|\frac1{n^s}-\frac1{(n+1)^s}\right| =\left|\frac1s\int_n^{n+1}\frac{du}{u^{s+1}}\right|\le\frac1{|s|n^{\sigma+1}}.$$ The series $$\sum_{n=1}^\infty a_n\left(\frac1{n^s}-\frac1{(n+1)^s}\right)$$ converges absolutely...

This argument works also for any Dirichlet L-series with non-trivial character.