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}

Decimal

0.75

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}

t an g e n t 1 - 4 x

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

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

\frac{d}{d x} (\frac{e ^{x}}{x})

x \to 0 + lim (arcsin (2 x - 1))

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