What is the sum from n=5 to infinity of 1/(n^2-1) ?

The sum from n=5 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=5 to infinity of 1/(n^2-1)

n = 5 \sum \infty \frac{1}{n ^{2} - 1}

converges

Solution steps

n = 5 \sum \infty \frac{1}{n ^{2} - 1}

Apply Series Integral Test:

converges

= converges

x \to 0 lim (6 x + 3)

\int a x + b x^{2} d x

x \to - 2 + lim (2 - x)

\frac{\partial}{\partial y} (2 e^{x} y)

y = x, (4, 2)

What is the sum from n=5 to infinity of 1/(n^2-1) ?
The sum from n=5 to infinity of 1/(n^2-1) is converges