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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{3}{4}

Solution steps

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 )}

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

Expand telescoping series:

\frac{3}{4}

= \frac{3}{4}

\frac{\partial}{\partial z} (ln (x + 2 y + 3 z))

x \to 3 lim (\frac{x ^{2} + x}{x - 3})

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

\int x csc^{2} (3 x) d x

\frac{\partial}{\partial y} (tan (7 + 2 x^{2} y^{3} z^{2}))

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