# How do you test the series Sigma (n+3)/(n(n+1)(n-2)) from n is [3,oo) for convergence?

Feb 9, 2017

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

is convergent.

#### Explanation:

Given:

${\sum}_{n = 3}^{\infty} {a}_{n} = {\sum}_{n = 3}^{\infty} \frac{n + 3}{n \left(n + 1\right) \left(n - 2\right)}$

we can use the direct comparison test, identifying a convergent series ${b}_{n}$ such that:

${a}_{n} < {b}_{n}$ for $n > N$

We can start from:

${a}_{n} = \frac{n + 3}{n \left(n + 1\right) \left(n - 2\right)}$

if we increase the numerator and decrease the denominator, the resulting quotient will be bigger:

${a}_{n} = \frac{n + 3}{n \left(n + 1\right) \left(n - 2\right)} = \frac{n - 2}{n \left(n + 1\right) \left(n - 2\right)} + \frac{5}{n \left(n + 1\right) \left(n - 2\right)} = \frac{1}{n \left(n + 1\right)} + \frac{5}{n \left(n + 1\right) \left(n - 2\right)}$

${a}_{n} < \frac{1}{n} ^ 2 + \frac{5}{n} ^ 2 = \frac{6}{n} ^ 2$

As ${\sum}_{n = 3}^{\infty} \frac{1}{n} ^ 2$ is convergent based on the p-series test, then also:

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

is convergent.