I've been struggling to prove this for a project as I'm not an expert in the field, so i decided to go back to basics.
The Dirichlet Eta Function is defined by:
$$ \eta (s) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{s}}, \text{ where } s= \sigma +it $$
The idea i have had is to use the fact that $n^{s} = n^{\sigma}n^{it} = n^{\sigma}(\cos(\ln(n)t)+i\sin(\ln(n)t))$
So we can re-write the sum as:
$$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{s}} = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{\sigma}(\cos(\ln(n)t)+i\sin(\ln(n)t))} = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{\sigma}} (\cos(\ln(n)t)-i\sin(\ln(n)t)) $$
So what we end up turning our original problem into two different real series:
$$ \eta (s) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{s}} = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{\sigma}}\cos(\ln(n)t) - i\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{\sigma}}\sin(\ln(n)t)) $$
So now my idea is to prove convergence for these two individual real series, which in turn will give me convergence for the complex series. Is there anything wrong with this approach?
