What is the sum from n=0 to infinity of 9/(n(n+3)) ?

The sum from n=0 to infinity of 9/(n(n+3)) is 11/2

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sum from n=0 to infinity of 9/(n(n+3))

n = 0 \sum \infty \frac{9}{n ( n + 3 )}

\frac{11}{2}

Solution steps

n = 1 \sum \infty \frac{9}{n ( n + 3 )}

Apply the constant multiplication rule:

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

= 9 \cdot n = 1 \sum \infty \frac{1}{n ( n + 3 )}

Take the partial fraction of

\frac{1}{n ( n + 3 )} : \frac{1}{3 n} - \frac{1}{3 ( n + 3 )}

= 9 \cdot n = 1 \sum \infty \frac{1}{3 n} - \frac{1}{3 ( n + 3 )}

Expand telescoping series:

\frac{11}{18}

Simplify

9 \cdot \frac{11}{18} : \frac{11}{2}

= \frac{11}{2}

\int (x^{2}) d x

\frac{d}{d x} ((2 x)^{2 x})

\frac{d}{d x} (4 x^{3} sec (7 x))

(x^{2} + 1) y^{^{'}} + 7 x (y - 1) = 0

(x - 2) y^{^{'}} - y = 2 (x - 2)^{3}, y (3) = 5

What is the sum from n=0 to infinity of 9/(n(n+3)) ?
The sum from n=0 to infinity of 9/(n(n+3)) is 11/2