Find the limet value of the residue $\displaystyle R_n$ when $\displaystyle n \rightarrow \infty$ in $\displaystyle z=1$ to the function $\displaystyle \displaystyle \frac{e^z}{(z+2)(z-1)^n}$.

I shuffled around a bit and used the geometric series and the expansion of $\displaystyle e^z$ and got this. $\displaystyle -e^1 \displaystyle \frac {\sum_{n=0}^{\infty} \frac{(z-1)^n} {n!} \sum_{n=0}^{\infty} (z-1)^n}{(z-1)^n}$

Now, I don't quite know how to simplify this expression... Ah well, here goes nothing!

$\displaystyle -e^1 \displaystyle \sum_{n=0}^{\infty} \frac{(z-1)^n (z-1)^n } {n! (z-1)^n}$ = $\displaystyle -e^1 \displaystyle \sum_{n=0}^{\infty} \frac{(z-1)^n } {n! }$

So, I'm not sure of how to simplify the expression with al the sums. And finally, I'm not sure of how to get further now...