Help me find my mistake when finding the exact value of the infinite sum $\sum_{n=0}^{\infty}\frac{e^{n-2}}{5^{n-1}}$ 
Finding the exact value of the infinite sum:
  $$\sum_{n=0}^{\infty}\frac{e^{n-2}}{5^{n-1}}$$

My Approach:
First term (a);
$$\frac{e^{-2}}{5^{-1}}=\frac{5}{e^2}$$
Second term:
$$\frac{e^{-1}}{5^{0}}=\frac{1}{e}$$
The difference (r):
$$r=\frac{e}{5}$$
Applying the geometric sum formula:
$$\frac{\frac{5}{e^2}}{1-\frac{e}{5}}=\frac{\frac{5}{e^2}}{\frac{5-e}{5}}$$
$$=\frac{25}{e^2(5-e)}$$
Unfortunately this is wrong, where have I gone wrong in this?

Your sum is correct.
One may recall a standard result concerning geometric series
$$
\sum_{n=0}^\infty r^n=\frac{r}{1-r},\qquad |r|<1.
$$ Applying it with $r=\dfrac{e}5\,\,\left(\left|\dfrac{e}5\right|<1\right)$, gives

$$
\sum_{n=0}^{\infty}\frac{e^{n-2}}{5^{n-1}}=\frac{e^{-2}}{5^{-1}}\sum_{n=0}^{\infty}\frac{e^{n}}{5^{n}}=\frac{e^{-2}}{5^{-1}}\frac1{1-e/5}=\frac{25}{e^2(5-e)}
$$ 

as you have found.
