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

1 Answer
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/(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.