What is the sum from n=9 to infinity of 1/(n(n-1)) ?

The sum from n=9 to infinity of 1/(n(n-1)) is 1/8

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sum from n=9 to infinity of 1/(n(n-1))

n = 9 \sum \infty \frac{1}{n ( n - 1 )}

\frac{1}{8}

Solution steps

n = 9 \sum \infty \frac{1}{n ( n - 1 )}

Apply Telescoping Series Test:

\frac{1}{8}

= \frac{1}{8}

x^{2} - 2 x

\int tan (\frac{x}{4}) d x

\frac{a}{1 - b x}

\frac{d y}{d x} - 5 y = x^{2}

\frac{7}{x}, - 1

What is the sum from n=9 to infinity of 1/(n(n-1)) ?
The sum from n=9 to infinity of 1/(n(n-1)) is 1/8