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

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

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

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

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

Solution steps

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

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

n = 0 \sum \infty \frac{3 ^{n}}{e ^{n}}

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

k = 2 \sum \infty (- 0.75)^{k}

n = 1 \sum \infty n^{- 0.3}

n = 1 \sum \infty \frac{sin ( nπ )}{n}

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