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?

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.. 
