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

Question

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?

Answer 1

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.