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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Integral Test:

converges

= 4 converges

= converges

\frac{7}{3 x - 1}

\frac{10}{s ^{3}}

x \to \infty lim (x^{2} + 12)

\frac{d}{d x} (\frac{π x}{4})

45 (1 - e^{- 625 x})

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