# Question #88d11

Apr 17, 2017

$\frac{11}{6}$

#### Explanation:

By Partial Fraction Decomposition,

$\frac{3}{n \left(n + 3\right)} = \frac{1}{n} - \frac{1}{n + 3}$

Let us find the partial sum:

${s}_{n} = \left(\frac{1}{1} - \cancel{\frac{1}{4}}\right) + \left(\frac{1}{2} - \cancel{\frac{1}{5}}\right) + \left(\frac{1}{3} - \cancel{\frac{1}{6}}\right) + \left(\cancel{\frac{1}{4}} - \cancel{\frac{1}{7}}\right) + \cdots + \left(\cancel{\frac{1}{n - 3}} - \cancel{\frac{1}{n}}\right) + \left(\cancel{\frac{1}{n - 2}} - \frac{1}{n + 1}\right) + \left(\cancel{\frac{1}{n - 1}} - \frac{1}{n + 2}\right) + \left(\cancel{\frac{1}{n}} - \frac{1}{n + 3}\right)$

By cleaning up the remaining terms,

$= \frac{11}{6} - \frac{1}{n + 1} - \frac{1}{n + 2} - \frac{1}{n + 3}$

Now, we can find the sum.

${\sum}_{n = 1}^{\infty} \frac{3}{n \left(n + 3\right)} = {\lim}_{n \to \infty} \left(\frac{11}{6} - \frac{1}{n + 1} - \frac{1}{n + 2} - \frac{1}{n + 3}\right) = \frac{11}{6} - 0 - 0 - 0 = \frac{11}{6}$

I hope that this was clear.