What is the sum from n=0 to infinity of e^{-n} ?

The sum from n=0 to infinity of e^{-n} is e/(e-1)

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sum from n=0 to infinity of e^{-n}

n = 0 \sum \infty e^{- n}

\frac{e}{e - 1}

Solution steps

n = 0 \sum \infty e^{- n}

Simplify

e^{- n} : \frac{1}{e ^{n}}

= n = 0 \sum \infty \frac{1}{e ^{n}}

Apply Series Geometric Test:

\frac{e}{e - 1}

= \frac{e}{e - 1}

\int 2 y^{2} d y

\int (\frac{x + 1}{( x - 3 ) ^{2}}) d x

\frac{\partial}{\partial z} (x yz)

\int (\frac{3}{x} - \frac{x}{3}) d x

(5, - 4), (7, - 4)

What is the sum from n=0 to infinity of e^{-n} ?
The sum from n=0 to infinity of e^{-n} is e/(e-1)