What is the sum from k=1 to infinity of e^{-k} ?

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

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sum from k=1 to infinity of e^{-k}

k = 1 \sum \infty e^{- k}

\frac{e}{e - 1} - 1

Decimal

0.58197 \dots

Solution steps

k = 1 \sum \infty e^{- k}

Simplify

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

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

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

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

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

= \frac{e}{e - 1} - \frac{1}{e ^{0}}

Simplify

= \frac{e}{e - 1} - 1

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