#sum_(n=1)^∞ (3k^2-3)/(k^5+1)#
The limit comparison test 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 = (3k^2-3)/(k^5+1)#
Let's think about the end behaviour of #a_n#.
For large #n#, the numerator #3k^2-3# acts like #3k^2#.
Also, for large #n#, the denominator #k^5+1# acts like #k^5#.
So, for large #n#, #a_n# acts like #(3k^2)/k^5 = 3/k^3#.
Let #b_n= 1/k^3#
Then #lim_(n→∞)(a_n/b_n) = lim_(n→∞)((3k^2-3)/(k^5+1))/(1/k^3)= lim_(n→∞)((3k^2-3)k^3)/(k^5+1) = lim_(n→∞)( (3k^5 +k^3)/(k^5+1)) = lim_(n→∞)( (3+1/k^2)/(1+1/k^5)) =3#
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/x^3# is convergent, so
#a_n = (3k^2-3)/(k^5+1)# is also convergent.
