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

Question

Christopher Allers · Accepted Answer

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.

Christopher Agresta · Answer

undefined To use the limit comparison test on the series $\sum_{n=1}^{\infty} \frac{n^2 - 5n}{n^3 + n + 1}$, follow these steps:

1. Choose a comparison series $b_n$ that is positive for all $n$ and whose convergence or divergence is known.

2. Take the limit of the ratio of the given series and the comparison series as $n$ approaches infinity:
   $$ \lim_{n \to \infty} \frac{a_n}{b_n} $$

3. If the limit is a finite positive number, both series either converge or diverge. If it's zero, the given series converges; if it's infinity, the given series diverges.

Choose a comparison series that simplifies the expression. For this series, let's use $b_n = \frac{1}{n}$, a p-series with $p = 1$, which is known to diverge.

Now, compute the limit:
$$ \lim_{n \to \infty} \frac{\frac{n^2 - 5n}{n^3 + n + 1}}{\frac{1}{n}} $$
$$ = \lim_{n \to \infty} \frac{n^2 - 5n}{n^3 + n + 1} \cdot n $$
$$ = \lim_{n \to \infty} \frac{n^3 - 5n^2}{n^3 + n + 1} $$

Now, divide each term by the highest power of $n$ in the denominator:
$$ = \lim_{n \to \infty} \frac{1 - \frac{5}{n}}{1 + \frac{1}{n^2} + \frac{1}{n^3}} $$

As $n$ approaches infinity, the terms with $1/n$ approach zero:
$$ = \frac{1}{1} = 1 $$

Since the limit is a finite positive number, the given series and the comparison series have the same behavior. Therefore, both series diverge.