What is the sum from n=3 to infinity of 1/(n(n-2)) ?

The sum from n=3 to infinity of 1/(n(n-2)) is 3/4

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sum from n=3 to infinity of 1/(n(n-2))

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

\frac{3}{4}

Solution steps

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

Take the partial fraction of

\frac{1}{n ( n - 2 )} : - \frac{1}{2 n} + \frac{1}{2 ( n - 2 )}

= n = 3 \sum \infty - \frac{1}{2 n} + \frac{1}{2 ( n - 2 )}

Expand telescoping series:

\frac{3}{4}

= \frac{3}{4}

n = 5 \sum \infty \frac{1}{n ^{2} ln ( n )}

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

n = 1 \sum \infty \frac{7}{n ^{\frac{2}{5}}}

n = 1 \sum \infty \frac{6 n}{3 n + 9}

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

What is the sum from n=3 to infinity of 1/(n(n-2)) ?
The sum from n=3 to infinity of 1/(n(n-2)) is 3/4