What is the sum from n=0 to infinity of 1/(n(n+2)) ?

The sum from n=0 to infinity of 1/(n(n+2)) is 3/4

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sum from n=0 to infinity of 1/(n(n+2))

n = 0 \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}

x \to 5 lim (x^{2} - 25)

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\int_{0}^{1} \frac{4 x + 6}{( x + 1 ) ^{2}} d x

\frac{d}{d x} (- x^{3} + 2 x^{2} - x)

\int \frac{3 x ^{2} - 2}{x ^{3} - 2 x - 8} d x

What is the sum from n=0 to infinity of 1/(n(n+2)) ?
The sum from n=0 to infinity of 1/(n(n+2)) is 3/4