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

Question

Emily Ammerman · Accepted Answer

#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/(n^4-1)# 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 = 1/n#. Let #b_n= 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= 1/n# is divergent, so #a_n = n^3/(n^4-1)# is divergent.

Charles Arocha · Answer

undefined To use the limit comparison test for the series $ \sum_{n=2}^{\infty} \frac{n^3}{n^4 - 1} $:

1. Identify a simpler series that you know converges or diverges.
2. Take the limit of the ratio of the given series to the simpler series as $ n $ approaches infinity.
3. If the limit is a positive finite number, then both series either converge or diverge together. If the limit is zero or infinite, the comparison test is inconclusive.

Let's choose the simpler series $ \sum_{n=2}^{\infty} \frac{1}{n} $ as our comparison series.

Now, take the limit:
$$ \lim_{n \to \infty} \frac{\frac{n^3}{n^4 - 1}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n^4}{n^4 - 1} $$

$$ = \lim_{n \to \infty} \frac{1}{1 - \frac{1}{n^4}} = 1 $$

Since the limit is a positive finite number, we conclude that both series either converge or diverge together. Since $ \sum_{n=2}^{\infty} \frac{1}{n} $ is a harmonic series which diverges, the original series $ \sum_{n=2}^{\infty} \frac{n^3}{n^4 - 1} $ also diverges by the limit comparison test.