What is the sum from k=1 to infinity of 6/(k(k+2)) ?

The sum from k=1 to infinity of 6/(k(k+2)) is 9/2

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sum from k=1 to infinity of 6/(k(k+2))

k = 1 \sum \infty \frac{6}{k ( k + 2 )}

\frac{9}{2}

Solution steps

k = 1 \sum \infty \frac{6}{k ( k + 2 )}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 6 \cdot 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 )}

= 6 \cdot k = 1 \sum \infty \frac{1}{2 k} - \frac{1}{2 ( k + 2 )}

Expand telescoping series:

\frac{3}{4}

Simplify

6 \cdot \frac{3}{4} : \frac{9}{2}

= \frac{9}{2}

k = 1 \sum \infty \frac{3}{k}

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

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

n = 1 \sum \infty \frac{4}{n}

n = 1 \sum \infty \frac{1}{5 n - 1}

What is the sum from k=1 to infinity of 6/(k(k+2)) ?
The sum from k=1 to infinity of 6/(k(k+2)) is 9/2