Question

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

Answer 1

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.

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I hope that this was helpful.