How do you use the limit comparison test to determine if #Sigma 1/(n(n^2+1))# from #[1,oo)# is convergent or divergent?

Question

Aurora Alvarado · Accepted Answer

The series:

#sum_(n=1)^oo 1/(n(n^2+1))#

is convergent.

The limit comparison test states that if we have two series with positive terms: #sum_(n=0)^oo a_n# and #sum_(n=0)^oo b_n# where the limit: #lim_(n->oo) a_n/b_n# exists and is finite, then if one series converges, the other is also convergent. Given: #a_n = 1/(n(n^2+1))# we can choose: #b_n = 1/(n^3)# that we know is convergent based on the p-series test and verify that: #lim _(n->oo) a_n/b_n = (1/(n(n^2+1)))/(1/(n^3)) =n^3/(n(n^2+1))= n^3/(n^3+n) = 1# So the series: #sum_(n=1)^oo 1/(n(n^2+1))# is convergent.

Scarlett Allison · Answer

undefined To use the limit comparison test to determine the convergence or divergence of the series $ \sum_{n=1}^{\infty} \frac{1}{n(n^2+1)} $, we compare it to a known series whose convergence behavior is understood.

Let's compare it to the series $ \sum_{n=1}^{\infty} \frac{1}{n^3} $.

We calculate the limit:

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

This limit is equal to 1, which means that the two series behave similarly. Since the series $ \sum_{n=1}^{\infty} \frac{1}{n^3} $ is known to be convergent (it is a p-series with $ p = 3 $ and $ p > 1 $), by the limit comparison test, the original series $ \sum_{n=1}^{\infty} \frac{1}{n(n^2+1)} $ is also convergent.