What is the value of this sum? 
Possible Duplicate:
Value of $\sum\limits_n x^n$ 

I am interested in finding what this sum converges to:
$$\sum_{n=0}^{\infty}e^{-n}=1+\frac{1}{e}+\frac{1}{e^2}+\frac{1}{e^3}+\cdots$$
Does a closed form exist?  If so, what is is?
 A: This is a classic geometric series. Letting
$$S=1+\frac{1}{e}+\frac{1}{e^2}+\frac{1}{e^3}+\cdots$$
we have
$$eS=e+1+\frac{1}{e}+\frac{1}{e^2}+\cdots$$
Taking the difference, we have
$$eS-S=S(e-1)=e+1+\frac{1}{e}+\frac{1}{e^2}+\cdots - \left(1+\frac{1}{e}+\frac{1}{e^2}+\frac{1}{e^3}+\cdots\right)$$
We see all the terms cancel each other out except for $e$.  Thus
$$S(e-1)=e\implies S = \frac{e}{e-1}$$
Thus the sum is equal to $\frac{e}{e-1}$.

More generally, we have
$$\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$$
for $|x| < 1$
A: Another way of seeing this result, using limits, is the following:
for any geometric sequence $\,a, ar,ar^2,\ldots , ar^n,\ldots\,$ , its $\,n\,$-th partial sum is$$S_n:=a+ar+...+ar^{n-1}=\sum_{k=0}^nar^k=\frac{a(1-r^n)}{1-r}$$which you can prove easily by induction on $\,n$ . From here, it follows that $$\sum_{k=0}^\infty ar^k=\lim_{n\to\infty}S_n=\lim_{n\to\infty}\frac{a(1-r^n)}{1-r}$$and the limit exists finitely (and thus the infinite series converges) iff $\,r^n\xrightarrow [n\to\infty]{} 0\Longleftrightarrow |r|<1\,$ , in which case$$\sum_{k=0}^\infty ar^n=\frac{a}{1-r}$$In your case, we have the infinites geometric series $$\sum_{k=0}^\infty \left(\frac{1}{e}\right)^n=\frac{1}{1-\frac{1}{e}}=\frac{e}{e-1}$$
