Geometric series sum $\sum_2^\infty e^{3-2n}$ $$\sum_2^\infty e^{3-2n}$$
The formulas for these things are so ambiguous I really have no clue on how to use them.
$$\frac {cr^M}{1-r}$$
$$\frac {1e^2}{1-\frac{1}{e}}$$
Is that a wrong application of the formula and why?
 A: Note $e^{3-2n}=e^3(e^{-2})^n$ so
$$\sum_{n=2}^\infty e^{3-2n}=e^3\sum_{n=2}^\infty (e^{-2})^n=e^3\left(\sum_{n=0}^\infty(e^{-2})^n-1-e^{-2}\right).$$
You can take it from here.
A: The specific formula you want is
$$\sum_{i=0}^\infty x^i = \frac{1}{1-x}.$$
There is no ambiguity there. But to apply this formula to the problem at hand, you first need to translate it so that you are summing from 0 to $\infty$, not from 2 to $\infty$. In this case, you can obtain
$$\sum_{i=2}^\infty e^{3-2i} = e^3\sum_{i=0}^\infty e^{-2(i+2)}$$
from the more usefully-shaped
$$e^3\sum_{i=0}^\infty e^{-2i}$$
by multiplying every term by $e^{-4}$. This is a common trick for geometric series - by just multiplying by powers of your radix, you can change which values you are summing over.
From there it seems like you've got the right idea, though your value for $r$ is slightly wrong. (Maybe this was a typo?)
A: The sum of a geometric series:
$$\sum_{k=0}^{\infty} ar^k = \frac{a}{1-r}$$
In your case:
$$r = e^{-2},\ a = e^3$$
Just remember to subtract the first two elements.. 
A: For me, it is most helpful to write out the first few terms of the series:
$$\sum_{n=2}^{\infty}e^{3-2n}=e^{-1}+e^{-3}+e^{-5}+\ldots$$
From this, it is easy to see that the initial term of the series is $a=e^{-1}$, and the common ratio is $r=e^{-2}$.  Using the formula for an infinite geometric series we have:
$$S=\frac{a}{1-r}=\frac{e^{-1}}{1-e^{-2}}=\frac{e}{e^2-1}$$
