What is the sum from n=2 to infinity of 1/(e^n) ?

The sum from n=2 to infinity of 1/(e^n) is (-e+1+e^2)/(e^2-e)-1

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sum from n=2 to infinity of 1/(e^n)

n = 2 \sum \infty \frac{1}{e ^{n}}

\frac{- e + 1 + e ^{2}}{e ^{2} - e} - 1

Solution steps

n = 2 \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}} - \frac{1}{e ^{1}}

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

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

Simplify

\frac{e}{e - 1} - \frac{1}{e ^{0}} - \frac{1}{e ^{1}} : \frac{- e + 1 + e ^{2}}{e ^{2} - e} - 1

= \frac{- e + 1 + e ^{2}}{e ^{2} - e} - 1

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What is the sum from n=2 to infinity of 1/(e^n) ?
The sum from n=2 to infinity of 1/(e^n) is (-e+1+e^2)/(e^2-e)-1