What is the sum from n=1 to infinity of 9/(n^2+3n) ?

The sum from n=1 to infinity of 9/(n^2+3n) is 11/2

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sum from n=1 to infinity of 9/(n^2+3n)

n = 1 \sum \infty \frac{9}{n ^{2} + 3 n}

\frac{11}{2}

Solution steps

n = 1 \sum \infty \frac{9}{n ^{2} + 3 n}

Apply the constant multiplication rule:

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

= 9 \cdot n = 1 \sum \infty \frac{1}{n ^{2} + 3 n}

Take the partial fraction of

\frac{1}{n ^{2} + 3 n} : \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}

n = 0 \sum \infty \frac{1}{n ^{- 2}}

n = 0 \sum \infty \frac{1}{9 - n}

n = 0 \sum \infty \frac{1}{1 - 7 n}

n = 0 \sum \infty \frac{1}{8 n + 4}

n = 0 \sum \infty \frac{n !}{9 ^{n}}

What is the sum from n=1 to infinity of 9/(n^2+3n) ?
The sum from n=1 to infinity of 9/(n^2+3n) is 11/2