What is the sum from n=1 to infinity of 5/(n(n+3)) ?

The sum from n=1 to infinity of 5/(n(n+3)) is 55/18

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 5/(n(n+3))

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

\frac{55}{18}

Solution steps

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

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty \frac{1}{n ( n + 3 )}

Take the partial fraction of

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

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

Expand telescoping series:

\frac{11}{18}

Simplify

5 \cdot \frac{11}{18} : \frac{55}{18}

= \frac{55}{18}

{1}

\int (\frac{x ^{2} - x + 5}{x}) d x

\frac{d y}{d x} = - y + 5

x \to - 7 lim (\frac{x + 3}{x + 7})

(4.4) (2.4)

What is the sum from n=1 to infinity of 5/(n(n+3)) ?
The sum from n=1 to infinity of 5/(n(n+3)) is 55/18