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人気のある微分積分 >

sum from n=1 to infinity of 1/(n(n+4))

解

n = 1 \sum \infty \frac{1}{n ( n + 4 )}

解

\frac{25}{48}

+1

十進法表記

0.52083 \dots

解答ステップ

n = 1 \sum \infty \frac{1}{n ( n + 4 )}

以下の部分分数を得る:

\frac{1}{n ( n + 4 )} : \frac{1}{4 n} - \frac{1}{4 ( n + 4 )}

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

畳み込み級数を拡張する:

\frac{25}{48}

= \frac{25}{48}

人気の例

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

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

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

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

sum from n=0 to infinity of (n+1)/(3^n)

n = 0 \sum \infty \frac{n + 1}{3 ^{n}}

sum from n=1 to infinity of (3n-2)/n

n = 1 \sum \infty \frac{3 n - 2}{n}

sum from n=0 to infinity of (-1/7)^n

n = 0 \sum \infty (- \frac{1}{7})^{n}