sum from n=70 to infinity of 3/(n^2+2n)

解答

n = 70 \sum \infty \frac{3}{n ^{2} + 2 n}

收敛

求解步骤

n = 70 \sum \infty \frac{3}{n ^{2} + 2 n}

使用常数相乘法则:

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

= 3 \cdot n = 70 \sum \infty \frac{1}{n ^{2} + 2 n}

\frac{1}{n ^{2} + 2 n} = \frac{n ^{2} - 2 n}{( n ^{2} ) ^{2} - ( 2 n ) ^{2}}

= 3 \cdot n = 70 \sum \infty \frac{n ^{2} - 2 n}{( n ^{2} ) ^{2} - ( 2 n ) ^{2}}

化简

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

= 3 \cdot n = 70 \sum \infty \frac{1}{n ( n + 2 )}

使用级数比较审敛法:

收敛

= 3 收敛

= 收敛

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

n = 1 \sum \infty (\frac{3}{5})^{n - 1}

n = 0 \sum \infty \frac{5}{5 - n}

n = 1 \sum \infty \frac{1}{n ^{3} + 6}

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