# How do you use the limit comparison test on the series sum_(n=1)^oo1/sqrt(n^3+1) ?

Sep 13, 2014

The Limit Comparison Test tells you that if
$0 < {\lim}_{n \to \infty} {a}_{n} / {b}_{n} < \infty$, with ${a}_{n} > 0 \mathmr{and} {b}_{n} > 0 ,$
then ${\sum}_{n = 1}^{\infty} {a}_{n}$ converges if and only if ${\sum}_{n = 1}^{\infty} {b}_{n}$ converges.

We want to find a series whose terms ${b}_{n}$ compare to our
${a}_{n} = \frac{1}{\sqrt{{n}^{3} + 1}}$; let's choose ${b}_{n} = \frac{1}{\sqrt{{n}^{3}}} = \frac{1}{{n}^{\frac{3}{2}}}$.

So ${\sum}_{n = 1}^{\infty} {b}_{n}$ converges by the Integral p-test; $p = \frac{3}{2} > 1.$
${\lim}_{n \to \infty} {a}_{n} / {b}_{n} = {\lim}_{n \to \infty} \frac{\sqrt{{n}^{3}}}{\sqrt{{n}^{3} + 1}} = \sqrt{{\lim}_{n \to \infty} \frac{{n}^{3}}{{n}^{3} + 1}} = 1$

Since $0 < 1 < \infty ,$ our series converges:
1. by the Limit Comparison Test, and
2. by the \ illustrious dansmath / ;-}