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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?

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    I see a vote for deletion of this question. Why?2012-06-22

2 Answers 2

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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 $|x| < 1$

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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 $\sum_{k=0}^\infty \left(\frac{1}{e}\right)^n=\frac{1}{1-\frac{1}{e}}=\frac{e}{e-1}$