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

The sum from n=1 to infinity of 3/(n^2-1) is 9/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

\frac{9}{4}

Decimal

2.25

Solution steps

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

Apply the constant multiplication rule:

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

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

Take the partial fraction of

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

= 3 \cdot n = 2 \sum \infty - \frac{1}{2 ( n + 1 )} + \frac{1}{2 ( n - 1 )}

Expand telescoping series:

\frac{3}{4}

Simplify

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

= \frac{9}{4}

What is the sum from n=1 to infinity of 3/(n^2-1) ?
The sum from n=1 to infinity of 3/(n^2-1) is 9/4