sum from n=4 to infinity of 1/n-1/(n+2)

解答

n = 4 \sum \infty \frac{1}{n} - \frac{1}{n + 2}

收敛

求解步骤

n = 4 \sum \infty \frac{1}{n} - \frac{1}{n + 2}

化简

\frac{1}{n} - \frac{1}{n + 2} : \frac{2}{n ( n + 2 )}

= n = 4 \sum \infty \frac{2}{n ( n + 2 )}

使用常数相乘法则:

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

= 2 \cdot n = 4 \sum \infty \frac{1}{n ( n + 2 )}

使用级数比较审敛法:

收敛

= 2 收敛

= 收敛

n = 0 \sum \infty \frac{n}{11 n + 1}

m = 2 \sum \infty \frac{3}{4 ^{m}}

n = 1 \sum \infty cos (\frac{3}{n ^{2}})

k = 4 \sum \infty \frac{1}{k ^{2} - 5 k + 6}

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