What is the sum from n=0 to infinity of 2/(4n^2-1) ?

The sum from n=0 to infinity of 2/(4n^2-1) is -1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

sum from n=0 to infinity of 2/(4n^2-1)

n = 0 \sum \infty \frac{2}{4 n ^{2} - 1}

- 1

Solution steps

n = 0 \sum \infty \frac{2}{4 n ^{2} - 1}

Apply the constant multiplication rule:

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

= 2 \cdot n = 0 \sum \infty \frac{1}{4 n ^{2} - 1}

Apply Telescoping Series Test:

- \frac{1}{2}

= 2 (- \frac{1}{2})

Simplify

2 (- \frac{1}{2}) : - 1

= - 1

What is the sum from n=0 to infinity of 2/(4n^2-1) ?
The sum from n=0 to infinity of 2/(4n^2-1) is -1