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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

converges

Solution steps

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

Apply Series Integral Test:

converges

= converges

\frac{s}{( s ^{2} + 9 ) ^{2}}

y = \frac{1}{4 - x}

\int \frac{( 4 x - 7 )}{x ^{2} - 6 x + 13} d x

- \frac{7}{x ^{2}}

y = 8 - x^{2}, y = x^{2}, x = - 3, x = 3

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