How do you use the limit comparison test on the series #sum_(n=1)^oo(n^2-5n)/(n^3+n+1)# ?

1 Answer
Oct 31, 2014

Let #a_n={n^2-5n}/{n^3+n+1}#.

By using the leading terms of the numerator and the denominator, we can construct

#b_n={n^2}/{n^3}=1/n#.

Remember that #sum_{n=1}^infty b_n# diverges since it is a harmonic series.

By Limit Comparison Test,

#lim_{n to infty}{a_n}/{b_n}=lim_{n to infty}{n^2-5n}/{n^3+n+1}cdot n/1 =lim_{n to infty}{n^3-5n^2}/{n^3+n+1}#

by dividing the numerator and the denominator by #n^3#,

#=lim_{n to infty}{1-5/n}/{1+1/n^2+1/n^3}={1-0}/{1+0+0}=1 < infty#,

which indicates that #sum_{n=1}^infty a_n# and #sum_{n=1}^infty b_n# are comparable.

Hence, #sum_{n=1}^infty{n^2-5n}/{n^3+n+1}# also diverges.


I hope that this was helpful.