# Limit Comparison Test for Convergence of an Infinite Series

## Key Questions

• 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.

• Let ${a}_{n} = \frac{{n}^{2} - 5 n}{{n}^{3} + n + 1}$.

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

${b}_{n} = \frac{{n}^{2}}{{n}^{3}} = \frac{1}{n}$.

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 ${n}^{3}$,

$= {\lim}_{n \to \infty} \frac{1 - \frac{5}{n}}{1 + \frac{1}{n} ^ 2 + \frac{1}{n} ^ 3} = \frac{1 - 0}{1 + 0 + 0} = 1 < \infty$,

which indicates that ${\sum}_{n = 1}^{\infty} {a}_{n}$ and ${\sum}_{n = 1}^{\infty} {b}_{n}$ are comparable.

Hence, ${\sum}_{n = 1}^{\infty} \frac{{n}^{2} - 5 n}{{n}^{3} + n + 1}$ also diverges.

I hope that this was helpful.