# How do you use the Nth term test on the infinite series sum_(n=1)^oo(n(n+2))/(n+3)^2 ?

Oct 10, 2014

Let ${a}_{n} = \frac{n \left(n + 2\right)}{{\left(n + 3\right)}^{2}} = \frac{{n}^{2} + 2 n}{{n}^{2} + 6 n + 9}$.

Since

${\lim}_{n \to \infty} {a}_{n} = {\lim}_{n \to \infty} \frac{{n}^{2} + 2 n}{{n}^{2} + 6 n + 9}$

by dividing the numerator and the denominator by ${n}^{2}$,

$= {\lim}_{n \to \infty} \frac{1 + \frac{2}{n}}{1 + \frac{6}{n} + \frac{9}{n} ^ 2} = \frac{1 + 0}{1 + 0 + 0} = 1 \ne 0$,

the series diverges by Nth Term (Divergence) Test.

I hope that this was helpful.