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人気のある微分積分 >

sum from k=0 to infinity of (-4/5)^k

解

k = 0 \sum \infty (- \frac{4}{5})^{k}

解

\frac{5}{9}

+1

十進法表記

0.55555 \dots

解答ステップ

k = 0 \sum \infty (- \frac{4}{5})^{k}

級数幾何テストを適用する:

\frac{5}{9}

= \frac{5}{9}

人気の例

sum from n=0 to infinity of 7(4/5)^n

n = 0 \sum \infty 7 (\frac{4}{5})^{n}

sum from n=1 to infinity of (n^2)/(8^n)

n = 1 \sum \infty \frac{n ^{2}}{8 ^{n}}

sum from n=1 to infinity of (9^n)/(n+1)

n = 1 \sum \infty \frac{9 ^{n}}{n + 1}

sum from n=0 to infinity of (10/11)^n

n = 0 \sum \infty (\frac{10}{11})^{n}

sum from i=0 to infinity of 1/(2^i)

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