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?
 A: Your method is true: by reindexing
$$\sum_2^\infty e^{3-2n}=\sum_{n=0}^\infty e^{-1-2n}=e^{-1}\frac{1}{1-e^{-2}}=\frac{1}{e-e^{-1}}$$
A: Here is how you advance
$$ \sum_2^\infty e^{3-2n}=e^3 \sum_2^\infty e^{-2n}=e^3\left( \sum_0^\infty e^{-2n}-1-e^{-2}  \right)=\dots.$$
You need the identity

$$ \sum_{n=0}^{\infty} x^n =\frac{1}{1-x}. $$

A: You should approach it this way (so that you avoid memorizing multiple formulas):
$$\sum_{n=2}^{\infty}e^{3-2n} = e^3\sum_{n=2}^{\infty}(e^{-2})^n = e^3\left(\sum_{n=0}^{\infty}(e^{-2})^n - (e^{-2})^0 - (e^{-2})^1\right).$$
Can you take it from here?
A: Recall that if $|x|<1$, then:

$$
\sum_{n=0}^\infty x^n = \dfrac{1}{1-x}
$$

With that in mind, observe that since $|e^{-2}|<1$, we have:
$$
\sum_{n=0}^\infty e^{-2n-1} = e^{-1} \sum_{n=0}^\infty (e^{-2})^{n} = \dfrac{e^{-1}}{1-e^{-2}}= \dfrac{e}{e^2-1}
$$
A: 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. 
