#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.
