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

Question

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

Charles Anctil · Accepted Answer

Let #a_n={n(n+2)}/{(n+3)^2}={n^2+2n}/{n^2+6n+9}#.

Since

#lim_{n to infty}a_n =lim_{n to infty}{n^2+2n}/{n^2+6n+9}#

by dividing the numerator and the denominator by #n^2#,

#=lim_{n to infty}{1+2/n}/{1+6/n+9/n^2}={1+0}/{1+0+0}=1 ne 0#,

the series diverges by Nth Term (Divergence) Test.

I hope that this was helpful.

Adam Adkins · Answer

undefined To use the Nth term test on the infinite series $ \sum_{n=1}^\infty \frac{n(n+2)}{(n+3)^2} $, follow these steps:

1. Examine the general term of the series, $ a_n = \frac{n(n+2)}{(n+3)^2} $.
  
2. Check if $ \lim_{n \to \infty} a_n \neq 0 $. If the limit is not equal to zero, the series diverges.

3. Calculate the limit $ \lim_{n \to \infty} a_n $. If the limit equals zero, the Nth term test is inconclusive, and other tests might be necessary to determine convergence or divergence.

In this case, $ a_n = \frac{n(n+2)}{(n+3)^2} $. Taking the limit as $ n $ approaches infinity:

$$ \lim_{n \to \infty} \frac{n(n+2)}{(n+3)^2} = \lim_{n \to \infty} \frac{n^2 + 2n}{n^2 + 6n + 9} = 1 $$

Since the limit is not equal to zero, the Nth term test is inconclusive for this series. You may need to use other convergence tests, such as the ratio test or the integral test, to determine the convergence or divergence of the series.