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

The sum from n=0 to infinity of 1/((n(n+2))) is 3/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from n=0 to infinity of 1/((n(n+2)))

n = 0 \sum \infty \frac{1}{( n ( n + 2 ) )}

\frac{3}{4}

Solution steps

n = 1 \sum \infty \frac{1}{n ( n + 2 )}

Take the partial fraction of

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

= n = 1 \sum \infty \frac{1}{2 n} - \frac{1}{2 ( n + 2 )}

Expand telescoping series:

\frac{3}{4}

= \frac{3}{4}

x + y = 4, x - y = 0, y + 3 x = 4

\frac{d}{d x} (\frac{4 x}{x ^{2} - 4})

\frac{d y}{d x} + y = e^{x}

f (x) = - \frac{6 x ^{2}}{( x ^{3} - 1 ) ^{2}}

\frac{d}{d x} ((x + 1)^{\frac{2}{3}} (2 x^{2} - 1)^{3})

What is the sum from n=0 to infinity of 1/((n(n+2))) ?
The sum from n=0 to infinity of 1/((n(n+2))) is 3/4