What is the sum from n=4 to infinity of 8/(n^2-1) ?

The sum from n=4 to infinity of 8/(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=4 to infinity of 8/(n^2-1)

n = 4 \sum \infty \frac{8}{n ^{2} - 1}

converges

Solution steps

n = 4 \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 = 4 \sum \infty \frac{1}{n ^{2} - 1}

Apply Series Integral Test:

converges

= 8 converges

= converges

2 x^{2}, 2 x + 6, - 1, 0

x y + 4 y = 10

x \to 2 lim (\frac{5 x ^{3}}{x + 3})

\int (\frac{1}{( x + 1 ) ^{2}}) d x

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

What is the sum from n=4 to infinity of 8/(n^2-1) ?
The sum from n=4 to infinity of 8/(n^2-1) is converges