The Limit Comparison Test for Positive Series of Real Numbers

# The Limit Comparison Test for Positive Series of Real Numbers

Recall from The Comparison Test for Positive Series of Real Numbers page that if $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ are both positive sequences of real numbers such that there exists an $N \in \mathbb{N}$ such that for all $n \geq N$ we have that $a_n \leq b_n$, then if $\displaystyle{\sum_{k=1}^{\infty} b_n}$ converges, we also have that $\displaystyle{\sum_{k=1}^{\infty} a_n}$ converges, and similarly, if $\displaystyle{\sum_{k=1}^{\infty} a_n}$ diverges then $\displaystyle{\sum_{k=1}^{\infty} b_n}$ diverges.

We will now look at another important comparison test that is sometimes easier to apply in certain situations.

Theorem 1: Let $\displaystyle{\sum_{n=1}^{\infty} a_n}$ and $\displaystyle{\sum_{n=1}^{\infty} b_n}$ be positive series of real numbers. Let $\displaystyle{\lim_{n \to \infty} \frac{a_n}{b_n} = L}$.a) If $0 < L < \infty$ then the series $\displaystyle{\sum_{n=1}^{\infty} a_n}$ and $\displaystyle{\sum_{n=1}^{\infty} b_n}$ either both converge or both diverge.b) If $L = 0$ and $\displaystyle{\sum_{n=1}^{\infty} b_n}$ converges then $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges.c) If $L = \infty$ and $\displaystyle{\sum_{n=1}^{\infty} b_n}$ diverges then $\displaystyle{\sum_{n=1}^{\infty} a_n}$ diverges. |

**Proof of a):**Suppose that $\displaystyle{\lim_{n \to \infty} \frac{a_n}{b_n} = L}$ where $0 < L < \infty$. Then there exists positive real numbers $m, M \in \mathbb{R}$ such that $m < L < M$.

- Now since $\displaystyle{\lim_{n \to \infty} \frac{a_n}{b_n} = L}$ we also have that there exists an $N \in \mathbb{N}$ such that if $n \geq N$ we have that:

\begin{align} \quad m < \frac{a_n}{b_n} < M \end{align}

- Or equivalently, for $n \geq N$ we see that:

\begin{align} \quad mb_n < a_n < Mb_n \end{align}

- Suppose that $\displaystyle{\sum_{k=1}^{\infty} a_n}$ converges. Then by the comparison test we see that the series $\displaystyle{\sum_{k=1}^{\infty} mb_n}$ also converges, so $\displaystyle{\sum_{k=1}^{\infty} b_n}$ converges.

- If instead $\displaystyle{\sum_{k=1}^{\infty} a_n}$ diverges then once again by the comparison test we see that the series $\displaystyle{\sum_{k=1}^{\infty} Mb_n}$ also diverges, so $\displaystyle{\sum_{k=1}^{\infty} b_n}$ diverges.

**Proof of b)**Suppose that $\displaystyle{\lim_{n \to \infty} \frac{a_n}{b_n} = 0}$. Then there exists a positive number $M \in \mathbb{R}$ such that for all $n \geq N$ we have that:

\begin{align} \quad 0 < \frac{a_n}{b_n} < M \end{align}

- So $0 < a_n < Mb_n$. If $\displaystyle{\sum_{k=1}^{\infty} b_n}$ converges then $\displaystyle{\sum_{k=1}^{\infty} Mb_n}$ converges, and by the comparison test we have that $\displaystyle{\sum_{k=1}^{\infty} a_n}$ also converges.

**Proof of c)**Suppose that $\displaystyle{\lim_{n \to \infty} \frac{a_n}{b_n} = \infty}$. Then there exists a positive number $m \in \mathbb{N}$ such that for all $n \geq N$ we have that:

\begin{align} \quad m < \frac{a_n}{b_n} < \infty \end{align}

- So $mb_n < a_n < \infty$. If $\displaystyle{\sum_{k=1}^{\infty} b_n}$ diverges then $\displaystyle{\sum_{k=1}^{\infty} mb_n}$ diverges, and by the comparison test we have that $\displaystyle{\sum_{k=1}^{\infty} a_n}$ also diverges. $\blacksquare$