Sum of $\sum_2^\infty e^{3-2n}$ $$\sum_2^\infty e^{3-2n}$$
I only have memorized the sum of at index zero. So I reindex
$$\sum_0^\infty e^{-2n-1}$$
This gives me the sum as
$$\frac{1}{1- \frac{1}{e} }$$
This is wrong. Why?

There is an $e^3$ factor that we can bring to the front. For the rest, we want 
$$\sum_{2}^\infty e^{-2n}.\tag{1}$$
It is nice to write  out (1) at greater length. It is 
$$\frac{1}{e^4}+\frac{1}{e^6}+\frac{1}{e^8}+\frac{1}{e^{10}}+\cdots.\tag(2)$$
Now if you know that when $|r|\lt 1$ then $a+ar+ar^2+ar^3+\cdots=\frac{a}{1-r}$, we are finished, for we have $a=\frac{1}{e^4}$ and $r=\frac{1}{e^2}$.
But if we don't want to remember that formula, take out a common factor of $\frac{1}{e^4}$. We get that (2) is equal to 
$$\frac{1}{e^4}\left(1+\frac{1}{e^2}+\frac{1}{e^4}+\frac{1}{e^6}+\cdots\right).$$
Now by the remembered formula, this is
$$\frac{1}{e^4}\cdot \frac{1}{1-\frac{1}{e^2}}.$$
Finally, remember about the $e^3$ we had temporarily forgotten about, and simplify. 
