So I'm trying to solve this practice problem I found on the internet. PracticeProblemLink
$$\displaystyle{ \sum_{n=0}^{\infty} {2e^{-n}} }$$
The part I'm stuck on is this: The general formula for the sum of the first n terms is Sn=[____]
I know that the common ratio is $$e^{-1}$$ and that the first term is $$2$$
And the formula for the nth term is $$S_n= \frac{a (1-r^n)}{(1-r)} $$
So I used this for this problem, getting the fraction $$ \frac{2(1-( e^{-n} )) }{(1-e^{-1})} $$
When I entered this into the problem it said I was incorrect. Am I missing something in the question?
Best Answer
Your formula is correct for the sum of the first $n$ terms (from $0$ to $n-1$): https://www.wolframalpha.com/input?i=Sum%5B2Exp%5B-k%5D%2C%7Bk%2C0%2Cn-1%7D%5D
Aside: it is a confusing choice to use $n$ as the summation index and then ask for the sum of the first $n$ terms.