The Lerch transcendent function is defined (see e.g. Wikipedia article) by $$ \Phi(z,s,a)=\sum_{n=0}^\infty\frac{z^n}{(n+a)^s}, $$ which is convergent for complex $z$ with $|z|<1$. It can be analytically continued to the whole complex $z$ plane, except a cut on $[1,+\infty)$, by the integral $$ \Phi(z,s,a)=\frac{1}{\Gamma(s)}\int_0^\infty \frac{t^{s-1}e^{-at}}{1-ze^{-t}}dt. $$
Question:
I would like to show that $\Phi(z,1,a)\to0$ as $z$ goes to complex infinity with the assumption that $0<a\le1$.
My approach:
Using the identity $$ \frac{1}{1-ze^{-t}}=1+\frac{ze^{-t}}{1-ze^{-t}} $$ the integral can be rewritten as $$ \Phi(z,1,a)=\int_0^\infty dt\,e^{-at}+ \int_0^\infty dt\,e^{-at}\frac{ze^{-t}}{1-ze^{-t}}. $$ The first term is just $a^{-1}$ and in the second term I can take the $z\to\infty$ limit under the integral to find (at leading order) $$ \Phi(z,1,a)\sim a^{-1}-a^{-1}=0,\qquad(z\to\infty). $$ This proves what I wanted but I am still skeptical whether this is correct; interestingly neither Maple nor Mathematica could calculate this.
Any ideas? Thanks