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

1 Answer
Sep 13, 2014

By Limit Comparison Test, we can conclude that
#sum_{n=1}^infty n/{2n^3+1}# converges.

Let us look at some details.

Let #a_n=n/{2n^3+1}#, and let #b_n=n/n^3=1/n^2#.
(Note: #b_n# was constructed by using the leading term of the numerator and that of the denominator ignoring the coefficients.)

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

by dividing the nuerator and denominator by #n^3#,
#=lim_{n to infty}1/{2+1/n^3}=1/2 < infty#

Since #sum_{n=1}^infty b_n=sum_{n=1}^infty1/n^2# is a convergent p-series with #p=2>1#,
#sum_{n=1}^infty a_n=sum_{n=1}^infty n/{2n^3+1}# also converges by Limit Comparison Test.