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

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

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

\frac{1}{2}

Solution steps

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

Apply Telescoping Series Test:

\frac{1}{2}

= \frac{1}{2}

- 2 x^{2} (x + 4)

\int_{- 2}^{2} (x^{3} - 3 x^{2} + 2) d x

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

\int \frac{50 x + 6}{( 7 x + 1 ) ( x - 1 )} d x

\int e^{4 x + 6} d x

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