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

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

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sum from n=4 to infinity of 1/(n(n+1))

n = 4 \sum \infty \frac{1}{n ( n + 1 )}

\frac{1}{4}

Solution steps

n = 4 \sum \infty \frac{1}{n ( n + 1 )}

Apply Telescoping Series Test:

\frac{1}{4}

= \frac{1}{4}

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

\int_{0}^{2} \frac{1}{x + 3} d x

cos (2 y)

\int \frac{1}{y ^{2} - 1} d y

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

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