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

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

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 2/(n(n+2))

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

\frac{3}{2}

Solution steps

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

Apply the constant multiplication rule:

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

= 2 \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 )}

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

Expand telescoping series:

\frac{3}{4}

Simplify

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

= \frac{3}{2}

\int \frac{sin ( x )}{cos ^{7} ( x )} d x

y^{^{'}} + 8 y - 6 = 0

\int r 1 + r^{2} d r

3 x^{2} e^{x} + e^{x} (x^{3} + 1)

\int - 4 x d x

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