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

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

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

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

\frac{3}{4}

Solution steps

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

Take the partial fraction of

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

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

Expand telescoping series:

\frac{3}{4}

= \frac{3}{4}

n = 1 \sum \infty \frac{n}{3 n ^{3} + 4}

n = 0 \sum \infty (0.03)^{n}

k = 1 \sum \infty (- \frac{6}{7})^{k}

k = 1 \sum \infty (\frac{2}{3})^{k + 2}

n = 1 \sum \infty \frac{4}{n ! + 4 ^{n}}

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