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

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

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sum from n=1 to infinity of (e)^{-n}

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

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

Solution steps

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

Simplify

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

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

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

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

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

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

Simplify

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

n = 2 \sum \infty \frac{1}{n} - \frac{1}{4 ^{n}}

n = 1 \sum \infty \frac{5 n}{5 n + 1}

k = 1 \sum \infty \frac{5}{k ( k + 2 )}

n = 1 \sum \infty \frac{3}{e ^{n} - 1}

n = 2 \sum \infty 2 e^{- 4 n}

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