I need to show that this complex series converges in C I, unfortunately, don't really know where to start. I know I can't use the ratio test. Should I try to split it up into a complex and a real part?



Translation: Show that for all $s\in\Bbb C\;$ such that $\operatorname{Re}s>0\;$ , the series $$h(x)=\sum_{k=0}^\infty\left[(4k+2)^{-s}-(4k+1)^{-s}\right]$$ converges in $\Bbb C$.
Thanks in advance for any suggestions.
 A: Some ideas for you:
First, observe that for any positive real $\;a\;$ , is we write $\;s=x+iy,\,x,y\in\Bbb R\;$, then
$$a^s=e^{(x+iy)\log a}=e^{x\log a}x^{iy\log a}\implies |x^a|=e^{x\log a}=a^x=a^{\text{Re}\,s}$$
Thus, we get that
$$=\left|\sum_{k=0}^\infty\left[(4k+2)^{-s}-(4k+1)^{-s}\right]\right|\le\sum_{k=0}^\infty\left[\frac1{|4k+2|^s}+\frac1{|4k+1|^s}\right]=$$
$$\sum_{k=0}^\infty\left[\frac1{(4k+2)^x}+\frac1{(4k+1)^x}\right]\;,\;\;s=x+iy\;,\;\;x,y,\in\Bbb R\,,\,\,x>0$$
and observe this last series is a positive one...Try to get it from here now.
A: One can use that $(4k+2)^{-s}-(4k+1)^{-s}=(-s)\int_{4k+1}^{4k+2}x^{-s-1}dx$ so
$|(4k+2)^{-s}-(4k+1)^{-s}| \le |s|\int_{4k+1}^{4k+2}|x^{-s-1}|dx \le |s|(4k+1)^{-\sigma-1}, \sigma=\Re s$
(since $|x^{-s-1}| \le (4k+1)^{-\sigma-1}, 4k+1 \le x \le 4k+2$)
This means that the series $\sum_{k=0}^\infty|\left[(4k+2)^{-s}-(4k+1)^{-s}\right]|$ is majorized by the absolutely convergent ($-\sigma-1 <-1$) series $\sum_{k \ge 0}|s|(4k+1)^{-\sigma-1}$ so we get absolute convergence for $h$
