sum from k=0 to infinity of e^{-k}

解答

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

\frac{e}{e - 1}

十进制

1.58197 \dots

求解步骤

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

化简

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

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

使用几何级数审敛法:

\frac{e}{e - 1}

= \frac{e}{e - 1}

n = 0 \sum \infty \frac{3}{5 - n}

n = 0 \sum \infty \frac{1}{( n ( n + 1 ) )}

n = 0 \sum \infty 1 + (- 1)^{n}

n = 0 \sum \infty ln (5 - n)

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