What is the sum from k=1 to infinity of 9/(k(k+2)) ?

The sum from k=1 to infinity of 9/(k(k+2)) is 27/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

sum from k=1 to infinity of 9/(k(k+2))

k = 1 \sum \infty \frac{9}{k ( k + 2 )}

\frac{27}{4}

Decimal

6.75

Solution steps

k = 1 \sum \infty \frac{9}{k ( k + 2 )}

Apply the constant multiplication rule:

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

= 9 \cdot k = 1 \sum \infty \frac{1}{k ( k + 2 )}

Take the partial fraction of

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

= 9 \cdot k = 1 \sum \infty \frac{1}{2 k} - \frac{1}{2 ( k + 2 )}

Expand telescoping series:

\frac{3}{4}

Simplify

9 \cdot \frac{3}{4} : \frac{27}{4}

= \frac{27}{4}

What is the sum from k=1 to infinity of 9/(k(k+2)) ?
The sum from k=1 to infinity of 9/(k(k+2)) is 27/4