This is part of a problem in Ahlfors, in the chapter on series and product developments.
$\eta(z) = 1^{-z} - 2^{-z} + 3^{-z} +\dots$ It is clear that the alternating series is convergent for positive reals. I am trying to prove convergence for the complex case for $z | Re(z) > 0$.
If $\eta_{m} = \sum_{l=1}^{m}\frac{(-1)^{l+1}}{l^z}$, I come to $\eta_{m}-\eta_n=\sum_{l=n+1}^{m}\frac{(-1)^{l+1}}{l^{\sigma}}e^{-it\log(l)}$ where in I substituted $z=\sigma+it$. Is it true that $\left|\sum_{l=n+1}^{m}\frac{(-1)^{l+1}}{l^{\sigma}}e^{-it\log(l)}\right| \leqslant \left|\sum_{l=n+1}^{m}\frac{(-1)^{l+1}}{l^{\sigma}}\,\right|?$ For if it were then the complex sum would be a contraction of the real sum for $\sigma > 0$ and convergence would be proved. Or are there better approaches?
Thanks in advance.