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

The sum from n=8 to infinity of 1/(n(n-1)) is 1/7

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

n = 8 \sum \infty \frac{1}{n ( n - 1 )}

\frac{1}{7}

Solution steps

n = 8 \sum \infty \frac{1}{n ( n - 1 )}

Apply Telescoping Series Test:

\frac{1}{7}

= \frac{1}{7}

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x \to 4 lim (\frac{x + 1}{x + 4})

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\frac{d y}{d x} = 6 e^{2 x - y}

What is the sum from n=8 to infinity of 1/(n(n-1)) ?
The sum from n=8 to infinity of 1/(n(n-1)) is 1/7