What is the sum from n=2 to infinity of 8/((n^2-1)) ?

The sum from n=2 to infinity of 8/((n^2-1)) is 6

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 8/((n^2-1))

n = 2 \sum \infty \frac{8}{( n ^{2} - 1 )}

6

Solution steps

n = 2 \sum \infty \frac{8}{n ^{2} - 1}

Apply the constant multiplication rule:

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

= 8 \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 )}

= 8 \cdot n = 2 \sum \infty - \frac{1}{2 ( n + 1 )} + \frac{1}{2 ( n - 1 )}

Expand telescoping series:

\frac{3}{4}

Simplify

8 \cdot \frac{3}{4} : 6

= 6

y = \frac{e ^{x}}{x}

\int (4 sin (x) - 1 + 8 x^{- 5}) d x

\int t^{2} - 4 t + 3 d t

\int sin (y) cos (x) d x

\int_{1}^{8} x ln (x) d x

What is the sum from n=2 to infinity of 8/((n^2-1)) ?
The sum from n=2 to infinity of 8/((n^2-1)) is 6