sum from k=0 to infinity of (-1/9)^k

解答

k = 0 \sum \infty (- \frac{1}{9})^{k}

\frac{9}{10}

十进制

0.9

求解步骤

k = 0 \sum \infty (- \frac{1}{9})^{k}

使用几何级数审敛法:

\frac{9}{10}

= \frac{9}{10}

n = 1 \sum \infty \frac{6}{n ^{2}}

x = 2 \sum \infty \frac{1}{x} - \frac{1}{x - 1}

k = 2 \sum \infty \frac{1}{( k - 1 ) ^{4}}

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

n = 2 \sum \infty \frac{1}{n ^{\frac{5}{3}}}