color(red)(sum_(n=2)^∞n^3/(n^4-1) " is divergent")∞∑n=2n3n4−1 is divergent.
sum_(n=2)^∞ n^3/(n^4-1)∞∑n=2n3n4−1
The limit comparison test (LCT) states that if a_nan and b_nbn 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.
