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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

1

Solution steps

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

Apply Telescoping Series Test:

1

= 1

\frac{d y}{d x}, 4 y^{2} - 5 y + 8 = 2 x^{2} - 3 x

\int 2 x (1 + x^{2})^{2} d x

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 e^{2 x}

\frac{d}{d x} (5 e^{- x} + e^{3 x})

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

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