# How do you use the limit comparison test for sum( n^3 / (n^4-1) )  from n=2 to n=oo?

Aug 11, 2015

color(red)(sum_(n=2)^∞n^3/(n^4-1) " is divergent").

sum_(n=2)^∞ n^3/(n^4-1)

The limit comparison test (LCT) states that if ${a}_{n}$ and ${b}_{n}$ are series with positive terms and if lim_(n→∞) (a_n)/(b_n) is positive and finite, then either both series converge or both diverge.

Let ${a}_{n} = {n}^{3} / \left({n}^{4} - 1\right)$

Let's think about the end behaviour of ${a}_{n}$.

For large $n$, the denominator ${n}^{4} - 1$ acts like ${n}^{4}$.

So, for large $n$, ${a}_{n}$ acts like ${n}^{3} / {n}^{4} = \frac{1}{n}$.

Let ${b}_{n} = \frac{1}{n}$

Then lim_(n→∞)(a_n/b_n) = lim_(n→∞)( (n^3/(n^4-1))/(1/n)) = lim_(n→∞)( (n^3×n)/(n^4-1)) = lim_(n→∞)( n^4/(n^4-1)) = lim_(n→∞)( 1/(1-1/n^4)) =1

The limit is both positive and finite, so either ${a}_{n}$ and ${b}_{n}$ are both divergent or both are convergent.

But ${b}_{n} = \frac{1}{n}$ is divergent, so

${a}_{n} = {n}^{3} / \left({n}^{4} - 1\right)$ is divergent.