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 1/2

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

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

\frac{1}{2}

Solution steps

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

Apply Telescoping Series Test:

\frac{1}{2}

= \frac{1}{2}

\int x^{\frac{- 2}{3}} x^{\frac{1}{3}} - 2 d x

\int \frac{d s}{s ^{2} ( s - 1 ) ^{2}} d x

\frac{d}{d x} (\frac{2}{x} - \frac{3 x}{3})

\frac{d}{d x} (\frac{x ^{3}}{4} + \frac{1}{3 x})

\int 2 d x

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 1/2