What is the sum from n=1 to infinity of 6/(n(n+2)) ?

The sum from n=1 to infinity of 6/(n(n+2)) is 9/2

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 6/(n(n+2))

n = 1 \sum \infty \frac{6}{n ( n + 2 )}

\frac{9}{2}

Solution steps

n = 1 \sum \infty \frac{6}{n ( n + 2 )}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty \frac{1}{n ( n + 2 )}

Take the partial fraction of

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

= 6 \cdot n = 1 \sum \infty \frac{1}{2 n} - \frac{1}{2 ( n + 2 )}

Expand telescoping series:

\frac{3}{4}

Simplify

6 \cdot \frac{3}{4} : \frac{9}{2}

= \frac{9}{2}

\frac{d y}{d x} = y^{2}

\int_{0}^{1} (x^{2} + 2) e^{- x} d x

\frac{d y}{d x} = 9 y e^{x + 5}

y^{^{'}} = x sin (2 y)

\int_{2}^{5} e^{2 x} d x

What is the sum from n=1 to infinity of 6/(n(n+2)) ?
The sum from n=1 to infinity of 6/(n(n+2)) is 9/2