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

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

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

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

\frac{e}{e - 1}

Solution steps

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

Simplify

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

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

Apply Series Geometric Test:

\frac{e}{e - 1}

= \frac{e}{e - 1}

n = 0 \sum \infty \frac{3}{5 - n}

n = 2 \sum \infty \frac{4}{n ln ( n )}

n = 1 \sum \infty \frac{6 ^{n}}{n}

n = 0 \sum \infty \frac{1}{( n ( n + 1 ) )}

n = 0 \sum \infty 1 + (- 1)^{n}

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