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

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Answer 1 · 2014-10-31T16:34:32

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.

I hope that this was helpful.

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