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

解答

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

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

十进制

0.21409 \dots

求解步骤

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}}

化简

\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}