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 1/3

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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}{3}

Solution steps

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

Apply Telescoping Series Test:

\frac{1}{3}

= \frac{1}{3}

n = 0 \sum \infty \frac{1}{( 1 - 3 n ) ^{3}}

n = 1 \sum \infty (\frac{e}{π})^{n + 1}

n = 1 \sum \infty \frac{1}{( n + 3 ) ^{3}}

k = 1 \sum \infty \frac{k ^{6}}{k !}

n = 7 \sum \infty \frac{1}{( 2 n - 1 ) ^{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 1/3