What is the sum from n=3 to infinity of 1/(n(n-1)) ?

The sum from n=3 to infinity of 1/(n(n-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=3 to infinity of 1/(n(n-1))

n = 3 \sum \infty \frac{1}{n ( n - 1 )}

\frac{1}{2}

Solution steps

n = 3 \sum \infty \frac{1}{n ( n - 1 )}

Apply Telescoping Series Test:

\frac{1}{2}

= \frac{1}{2}

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

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

\int \frac{1}{x 1 - x} d x

x^{6} (5.1)^{x}

x \to 0 lim ((\frac{1}{x ^{2}})^{x})

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