sum from k=4 to infinity of 6/k-6/(k+1)

解答

k = 4 \sum \infty \frac{6}{k} - \frac{6}{k + 1}

\frac{3}{2}

十进制

1.5

求解步骤

k = 4 \sum \infty \frac{6}{k} - \frac{6}{k + 1}

乘开

\frac{6}{k} - \frac{6}{k + 1} : \frac{6}{k ^{2} + k}

= k = 4 \sum \infty \frac{6}{k ^{2} + k}

使用常数相乘法则:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 6 \cdot k = 4 \sum \infty \frac{1}{k ^{2} + k}

使用裂项级数审敛法:

\frac{1}{4}

= 6 \cdot \frac{1}{4}

化简

6 \cdot \frac{1}{4} : \frac{3}{2}

= \frac{3}{2}

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

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

n = 0 \sum \infty 4 (\frac{8}{9})^{n}

n = 1 \sum \infty cos (n)

n = 0 \sum \infty n^{2} - 2