What is the sum from n=1 to infinity of 1/(n^2+3n) ?

The sum from n=1 to infinity of 1/(n^2+3n) is 11/18

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{11}{18}

Solution steps

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

Take the partial fraction of

\frac{1}{n ^{2} + 3 n} : \frac{1}{3 n} - \frac{1}{3 ( n + 3 )}

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

Expand telescoping series:

\frac{11}{18}

= \frac{11}{18}

f (x) = 270

y = x^{3} - 4 x^{2} + 3 x, y = x^{2} - x

\frac{d y}{d x} = x + 6 y

\frac{\partial}{\partial x} (cos (x + 5 y))

(2 x + 5) (- x + 2)

What is the sum from n=1 to infinity of 1/(n^2+3n) ?
The sum from n=1 to infinity of 1/(n^2+3n) is 11/18