Why does $n = \sum_{k=0}^\infty \frac{1}{2^k\;e^{k-2}} \implies n = \frac{2\;e^3}{2\;e-1}$? Consider $$n = \sum_{k=0}^\infty \frac{1}{2^k\;e^{k-2}}$$
Why does
$$n = \frac{2\;e^3}{2\;e-1}\;\;?$$

HINT:
$$n = \sum_{k=0}^\infty \frac{1}{2^k\;e^{k-2}}=e^2\sum_{k=0}^\infty\frac1{(2e)^k}$$
Observe that $\sum_{k=0}^\infty\frac1{(2e)^k}$ is an Infinite Geometric Series with the first term $\frac1{(2e)^0}=1$ and common ratio $\frac1{2e}$ which lies $\in(0,1)$ 
