The sum of geometric series $e^{k-1}/\pi^{k+1}$ 
Let $T_n=\sum _{k=1}^{n}\dfrac{e^{k-1}}{\pi ^{k+1}}$ calculate the $\lim_{n\to\infty}T_n$

Note $T_n$ is a geometric series:
\begin{align*}
T_n&=\sum _{k=1}^{n \:}\dfrac{e^{k-1}}{\pi ^{k+1}}\\
&= \pi^{-2}.\sum _{k=1}^{n \:}\dfrac{e^{k-1}}{\pi ^{k-1}}\\
&= \pi^{-2}.\sum _{k=1}^{n \:}\left(\dfrac{e}{\pi }\right)^{k-1}\\
&= \pi^{-2}.\sum _{k=0}^{n-1 \:}\left(\dfrac{e}{\pi }\right)^{k}(\text{change of index})\\
&= \pi^{-2}.\dfrac{1-\left(\dfrac{e}{\pi }\right)^{n}}{1-\left(\dfrac{e}{\pi }\right)}(\text{Geometric series})\\
&= \pi^{-1-n}.\dfrac{\pi^n-e^n }{\pi-e }(\text{Geometric series})\\
\text{then the limit of $T_n$ }\\
\lim_{n\to\infty}T_n&=\lim_{n\to\infty} \pi^{-1-n}.\dfrac{\pi^n-e^n }{\pi-e } ??
\end{align*}
I'm stuck in limit.

You just need to take the limit one step earlier. That is, find 
$$\lim_{n\to\infty}\pi^{-2}\frac{1-\left(\dfrac{e}{\pi}\right)^n}{1-\dfrac{e}{\pi}}$$
by using the fact that if $|x| < 1$ then $\displaystyle\lim_{n \to \infty}x^n = 0$.
