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

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

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

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

\frac{15}{4}

Solution steps

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

Apply the constant multiplication rule:

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

= 5 \cdot 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 )}

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

Expand telescoping series:

\frac{3}{4}

Simplify

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

= \frac{15}{4}

\frac{d}{d x} (\frac{x x + 1}{x + 1})

y^{^{'}} + 6 y = e^{t}

\int \frac{5}{3 - 2 cos ( x )} d x

x \to 3 lim (4 π)

\frac{\partial}{\partial x} (x cos (\frac{y}{z}))

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