$$\sum_2^\infty e^{3-2n}$$
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?
Calculus – Geometric Series Sum $\sum_2^\infty e^{3-2n}$
calculussequences-and-series
Best Answer
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.