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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{2}

Solution steps

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

Apply Telescoping Series Test:

\frac{1}{2}

= \frac{1}{2}

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

n = 1 \sum \infty cos (6)

n = 1 \sum \infty \frac{n}{5 n ^{2} - 4}

n = 0 \sum \infty tanh (n)

n = 0 \sum \infty 8

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