The Limit Comparison Test for Positive Series

We will now look at another comparison test similar to The Comparison Test for Positive Series known as The Limit Comparison Test for Positive Series.

Theorem 1 (The Limit Comparison Test for Positive Series): Let $\{ a_n \}$ and $\{ b_n \}$ be positive sequences and suppose that $\lim_{n \to \infty} \frac{a_n}{b_n} = L$ for some $L$. Then: a) If $0 < L < \infty$ then either both of the series $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} b_n$ both converge or both diverge. b) If $L = 0$ and $\sum_{n=1}^{\infty} b_n$ converges then $\sum_{n=1}^{\infty} a_n$ converges. If $L = 0$ and $\sum_{n=1}^{\infty} b_n$ diverges, then the test is inconclusive. c) If $L = \infty$ and $\sum_{n=1}^{\infty} b_n$ diverges then $\sum_{n=1}^{\infty} a_n$ diverges. If $L = \infty$ and $\sum_{n=1}^{\infty} b_n$ converges, then the test is inconclusive.

The idea with the limit comparison test is to take some series that whose convergence/divergence we don't know, then take another series that we do know is convergent/divergent.

• Proof of a): Let $\lim_{n \to \infty} \frac{a_n}{b_n} = L$ such that $0 < L < \infty$. There exists two positive numbers $m, M \in \mathbb{R}$ such that $m < L < M$. Now for some $N \in \mathbb{N}$ if $n ≥ N$ then $m < \frac{a_n}{b_n} < M$ or rather $mb_n < a_n < Mb_n$ for $n ≥ N$.

• Now notice that if $\sum_{n=1}^{\infty} b_n$ converges then $\sum_{n=1}^{\infty} Mb_n$ also converges and by comparison since $a_n < Mb_n$ then $\sum_{n=1}^{\infty} a_n$ must also converge.

• Similarly note that if $\sum_{n=1}^{\infty} b_n$ diverges then $\sum_{n=1}^{\infty} mb_n$ also diverges and by comparison since $mb_n < a_n$ then $\sum_{n=1}^{\infty} a_n$ must also diverge.

• Lastly, we note that if $L = 0$ then we cannot set up the same inequality used above, so this test in inconclusive. $\blacksquare$

We will now look at some examples applying the limit comparison test for positive series.

Example 1

Determine whether the series $\sum_{n=1}^{\infty} \frac{n^2 + 2}{n^4 + n + 10}$ converges or diverges using the limit comparison test.

We first note that this series is positive for $n ≥ 1$ and that the terms in this series behave like $\frac{1}{n^2}$ for $n$ sufficiently large, use $\sum_{n=1}^{\infty} \frac{1}{n^2}$ to compare with.

(1)

\begin{align} \lim_{n \to \infty} \frac{\frac{n^2 + 2}{n^4 + n + 10}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{n^4 + 2n^2}{n^4 + n + 10} = \lim_{n \to \infty} \frac{1 + \frac{2}{n^2}}{1 + \frac{1}{n^3} + \frac{10}{n^4}} = 1 \end{align}

We note that by the p-Series test that $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges and thus by the limit comparison test then $\sum_{n=1}^{\infty} \frac{n^2 + 2}{n^4 + n + 10}$ must also converge.

Example 2

Determine whether the series $\sum_{n=1}^{\infty} \frac{n^4 + n^3 + n^2 + n + 1}{n^6 - n + 2}$ converges or diverges using the limit comparison test.

We note that this series is positive for $n ≥ 1$ and that the terms in this series behave like $\frac{1}{n^2}$ once again for $n$ sufficiently large, so we will use $\sum_{n=1}^{\infty} \frac{1}{n^2}$ to compare with.

(2)

\begin{align} \quad \lim_{n \to \infty} \frac{\frac{n^4 + n^3 + n^2 + n + 1}{n^6 - n + 2}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{n^6 + n^5 + n^4 + n^3 + n^2}{n^6 - n + 2} = \lim_{n \to \infty} \frac{1 + \frac{1}{n} + \frac{1}{n^2} + \frac{1}{n^3} + \frac{1}{n^4}}{1 - \frac{1}{n^5} + \frac{2}{n^6}} = 1 \end{align}

By the p-Series test we know that $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges and thus by the limit comparison test then $\sum_{n=1}^{\infty} \frac{n^4 + n^3 + n^2 + n + 1}{n^6 - n + 2}$ must also converge.