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

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

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

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

\frac{3}{4}

Solution steps

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

Expand telescoping series:

\frac{3}{4}

= \frac{3}{4}

1 - 4 x

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

\frac{d ^{2}}{d x ^{2}} (ln (1 + e^{x}))

\frac{\partial}{\partial x} (7 x + 8 y^{2})

\frac{s + 11}{( s - 10 ) ( s + 4 )}

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