uniform convergence alternating series in Titsmarsch book on zeta function page 17 he said that this series 
 $\sum_{m=1}^\infty (-1)^m \{(m+1)^s - m^s\}$ is uniformly convergent on $0 < \Re(s) < 1$
for real $s$ i can see how it is but for complex $s$ i'm not sure can someone please help? thanks    
 A: Let $s\in \mathbb{C}$ and let $f(s)$ be given by
$$f(s)=\sum_{m=1}^\infty (-1)^{m}\left((m+1)^s-m^s \right) \tag 1$$
In THIS ANSWER, I used a Generalized Dirichlet Test (See here) to show that the alternating series representation $\eta(s)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}$ for the Dirichlet Eta function converges for $\text{Re}(s)>0$.  
Noting that $\left((m+1)^s-m^s \right)=\frac s{m^{1-s}} +O\left(\frac1{m^{2-s}}\right)$, it is easily seen that the series on the right-hand side of $(1)$ converges for $0<\text{Re}(s)<1$. 
Hence, by pairing successive even and odd terms, we can write
$$f(s)=\sum_{m=1}^\infty \left((2m+1)^s-2(2m)^s+(2m-1)^s\right)$$
Now, note that $\left((2m+1)^s-2(2m)^s+(2m-1)^s\right)=\frac{2^ss(s-1)}{4m^{2-s}}+O\left(\frac1{m^{4-s}}\right)$.  Furthermore, for any $\delta>0$ and $0<\text{Re}(s)<1-\delta$ we have the estimates 
$$\begin{align}
\left|\sum_{m=1}^\infty \frac1{m^{2-s}}\right|&\le\sum_{m=1}^\infty \frac1{\left|m^{2-s}\right|}\\\\
&= \sum_{m=1}^\infty \frac{1}{m^{2-\text{Re}(s)}}\\\\
&\le \sum_{m=1}^\infty \frac{1}{m^{1+\delta}}
\end{align}$$
This shows that the series in $(1)$ converges uniformly on all compact subsets contained on $0<\text{Re}(s)<1$.
