Show that $\sum ^{\infty}_{n=1} e^{-n} \sin nz$ is analytic in the region $A=\{z|-1<Im \;z<1\}$
Idea:
Let $D$ be a closed disk in $A$, let $\delta$ be its distance from the boundary $Im \;z = \pm1$
take $z=x+iy \in D$
i have to show that $| e^{-n} \sin nz| \le e^{-n\delta}$ ..how to this prove
Recall that the uniform limit of analytic functions is analytic. As the finite sum of analytic functions, each partial sum is analytic. So, you need to show uniform convergence, which is perhaps what you were doing.
Computing the modulus of the summand, $$ \left| e^{-n}\frac{e^{niz}-e^{-niz}}{2i} \right|\leq e^{-n}|e^{niz}| $$ now write $z=x+iy$, and you find $$ e^{-n}|e^{nix-ny}|=e^{-n}e^{-ny} $$ now as long as we can get some separation from the left hand point of the strip, i.e. if we can take $$ -1+\delta<\Im(z) $$ for some $\delta>0$, then the series converges uniformly by the Weirstrass M test and since $$ e^{-n}e^{-ny}\leq e^{-n}e^{-n(-1+\delta)}=e^{-n\delta} $$ which is summable.