sum from k=1 to infinity of 2^{-k}

解答

k = 1 \sum \infty 2^{- k}

1

求解步骤

k = 1 \sum \infty 2^{- k}

化简

2^{- k} : \frac{1}{2 ^{k}}

= k = 1 \sum \infty \frac{1}{2 ^{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}{2 ^{k}} - \frac{1}{2 ^{0}}

k = 0 \sum \infty \frac{1}{2 ^{k}} = 2

= 2 - \frac{1}{2 ^{0}}

化简

= 1

k = 5 \sum \infty (\frac{e}{π})^{k - 1}

k = 4 \sum \infty \frac{1}{( k - 1 ) ^{4}}

n = 0 \sum \infty 6 (\frac{2}{3})^{n}

n = 0 \sum \infty 2 (\frac{1}{2})^{n}

n = 1 \sum \infty (\frac{3}{4})^{n - 1}