Question

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

Answer 1

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.