# 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)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)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.