sum from n=1 to infinity of 1/(n^2+5n)

解答

n = 1 \sum \infty \frac{1}{n ^{2} + 5 n}

收敛

求解步骤

n = 1 \sum \infty \frac{1}{n ^{2} + 5 n}

将

\frac{1}{n ^{2} + 5 n}

用部份分式展开:

\frac{1}{5 n} - \frac{1}{5 ( n + 5 )}

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

乘以

\frac{1}{5 n} - \frac{1}{5 ( n + 5 )}

的共轭根式

: \frac{\frac{1}{25 n ^{2}} - \frac{1}{25 n ^{2} + 250 n + 625}}{\frac{1}{5 n} + \frac{1}{5 ( n + 5 )}}

= n = 1 \sum \infty \frac{\frac{1}{25 n ^{2}} - \frac{1}{25 n ^{2} + 250 n + 625}}{\frac{1}{5 n} + \frac{1}{5 ( n + 5 )}}

化简

\frac{\frac{1}{25 n ^{2}} - \frac{1}{25 n ^{2} + 250 n + 625}}{\frac{1}{5 n} + \frac{1}{5 ( n + 5 )}} : \frac{1}{n ( n + 5 )}

= n = 1 \sum \infty \frac{1}{n ( n + 5 )}

使用级数比较审敛法:

收敛

= 收敛

x \to - 4 lim (x^{2} - 16)

\frac{d}{d x} (\frac{x ^{\frac{1}{2}}}{3 x})

a re a tan (3 x), 2 sin (3 x), [- \frac{π}{9}, \frac{π}{9}]

d er i v a t i v e y = \frac{x - 1}{x + 1}

\int \frac{( 4 x - 11 )}{( x ^{3} - 9 x ^{2} )} d x