Limit Comparison Test for Convergence of an Infinite Series

Key Questions

How do I know when to use limit comparison test vs the direct comparison test?

Well, I would try to see if I can directly compare first; however, it might not be easy when its expression is complicated. The benefit of the limit comparison test is that we can compare series without verifying the inequality we need in order to apply the direct comparison test, of course, at the cost of having to evaluate the limit.

Wataru · 3 · Aug 27 2014
How do you use the limit comparison test on the series $\infty \sum n = 1 \frac{n^{2} - 5 n}{n^{3} + n + 1}$ $sum_(n=1)^oo(n^2-5n)/(n^3+n+1)$ ?

Let $a_{n} = \frac{n^{2} - 5 n}{n^{3} + n + 1}$ $a_n={n^2-5n}/{n^3+n+1}$ .

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

$b_{n} = \frac{n^{2}}{n^{3}} = \frac{1}{n}$ $b_n={n^2}/{n^3}=1/n$ .

Remember that $\infty \sum n = 1 b_{n}$ $sum_{n=1}^infty b_n$ 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.

Wataru · 1 · Oct 31 2014