# proving analyticity of a complex series.

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?

Being not very rigorous, it's evident that the vectors $e^{-it log(l)}$ do not differ significantly for large values of l and if m-n is also not too large. In those cases the inequality may hold (locally)?? – vnd Dec 30 '12 at 19:34
The quickest road might be to note that, when $t=\Re(z)$ is positive, $n^{-z}\to0$ when $n\to\infty$, and to consider the sums of two consecutive terms $$s_n=(2n-1)^{-z}-(2n)^{-z},$$ for every $n\geqslant1$. Then, $$s_n=z\int_{2n-1}^{2n}x^{-z-1}\mathrm dx,$$ hence $$∣s_n∣\,\leqslant\,∣z∣\cdot\int_{2n-1}^{2n}x^{-t-1}\mathrm dx\,\leqslant\, ∣z∣\cdot(2n-1)^{-t-1}.$$ In particular, $s_n=O(n^{-t-1})$ hence the series $\sum\limits_ns_n$ converges (absolutely) as soon as $t\gt0$, a fact which implies the (simple) convergence of $\eta(z)$.