What is the sum from k=1 to infinity of 5/(k(k+2)) ?

The sum from k=1 to infinity of 5/(k(k+2)) is 15/4

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sum from k=1 to infinity of 5/(k(k+2))

k = 1 \sum \infty \frac{5}{k ( k + 2 )}

\frac{15}{4}

Solution steps

k = 1 \sum \infty \frac{5}{k ( k + 2 )}

Apply the constant multiplication rule:

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

= 5 \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 )}

= 5 \cdot k = 1 \sum \infty \frac{1}{2 k} - \frac{1}{2 ( k + 2 )}

Expand telescoping series:

\frac{3}{4}

Simplify

5 \cdot \frac{3}{4} : \frac{15}{4}

= \frac{15}{4}

n = 1 \sum \infty \frac{3}{e ^{n} - 1}

n = 2 \sum \infty 2 e^{- 4 n}

n = 1 \sum \infty \frac{i}{5 ^{n}}

n = 1 \sum \infty \frac{1}{cos ( \frac{1}{n} )}

n = 0 \sum \infty (n + 8)^{2}

What is the sum from k=1 to infinity of 5/(k(k+2)) ?
The sum from k=1 to infinity of 5/(k(k+2)) is 15/4