The Ratio Test for Sequence Convergence

# The Ratio Test for Sequence Convergence

We will now look at a useful theorem that we can apply in order to determine whether a sequence of positive real numbers converges. Before we do so, we must first prove the following lemma.

 Lemma 1: Let $(a_n)$ be a sequence of positive real numbers, and $(b_n)$ be a sequence of real numbers, and let $K > 0$. If $\lim_{n \to \infty} a_n = 0$ and for some $N_1 \in \mathbb{N}$ we have that $\forall n ≥ N_1$ that $\mid b_n - L \mid ≤ K a_n$, then $\lim_{n \to \infty} b_n = L$.
• Proof: Let $(a_n)$ be a sequence of positive real numbers and $(b_n)$ be a sequence of real numbers and let $K > 0$. Suppose that $\lim_{n \to \infty} a_n = 0$ and that for some $N_1 \in \mathbb{N}$ that if $n ≥ N_1$ then $\mid b_n - L \mid < Ka_n$.
• Now let $\epsilon > 0$ be given. Since $\lim_{n \to \infty} a_n = 0$ then for $\epsilon_0 = \frac{\epsilon}{K} > 0$ then $\exists N_2 \in \mathbb{N}$ such that if $n ≥ N_2$ then $\mid a_n \mid = a_n < \epsilon_0 = \frac{\epsilon}{K}$, which implies that $Ka_n < \epsilon$
• Now let $N = \mathrm{max} \{ N_1, N_2 \}$. Then we have that:
(1)
\begin{align} \mid b_n - L \mid ≤ Ka_n < \epsilon \end{align}
• Therefore $\lim_{n \to \infty} b_n = L$. $\blacksquare$

We are now ready to look at the Ratio Test for positive sequence convergence.

 Theorem 1 (The Ratio Test for Sequences): If $(a_n)$ is a sequence of positive real numbers such that $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L$ and if $L < 1$, then $(a_n)$ converges and $\lim_{n \to \infty} a_n = 0$.

For example, consider the sequence $(a_n) = \left ( \frac{2^n}{3^n} \right )$. Applying the ratio test we get that $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{2^{n+1}}{3^{n+1}} \cdot \frac{3^n}{2^n} = \lim_{n \to \infty} \frac{2}{3} < 1$. Since $(a_n)$ is a positive sequence, we deduce that by the ratio test, this sequence converges, namely to $0$.

We will now look at the proof of this theorem.

• Proof: Let $(a_n)$ be a sequence of positive real numbers such that $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L$. Since $(a_n)$ is a sequence of positive real numbers, then we have that $0 ≤ L$. Suppose also that $L < 1$. Let $r$ be such that $L < r < 1$ (we know that such a number $r \in \mathbb{R}$ exists by the The Density of Real Numbers Theorem as both rational and irrational numbers exist between $L$ and $1$).
• Since $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L$, then for $\epsilon = r - L > 0$, $\exists N \in \mathbb{N}$ such that if $n ≥ N$ then $\biggr \rvert \frac{a_{n+1}}{a_n} - L \biggr \rvert < \epsilon = r - L$, which implies that:
(2)
\begin{align} - \epsilon < \frac{a_{n+1}}{a_n} - L < \epsilon \\ L - \epsilon < \frac{a_{n+1}}{a_n} < L + \epsilon \\ \frac{a_{n+1}}{a_n} < L + (r - L) = r \end{align}
• Therefore for $n ≥ N$ we have that $a_{n+1} < a_n r$. But also notice that $a_nr < a_{n-1}r^2$, and $a_{n-1}r^2 < a_{n-2}r^3$, …, $a_{N+1} r^{n - N} < a_N r^{n-N+1}$. We thus obtain the following inequality:
(3)
\begin{equation} a_{n+1} < a_nr < a_{n-1} r^2 < a_{n-2} r^3 < ... < a_{N+1} r^{n-N} < a_N r^{n - N + 1} \end{equation}
• Now let $K = \frac{a_N}{r^N}$. Therefore $a_N r^{n-N+1} = K r^{n+1}$. Thus for $n ≥ N$ we have that $a_nr < Kr^{n+1}$ or rather $a_n < Kr^n$. Now since $0 < r < 1$, we have that $\lim_{n \to \infty} r^n = 0$. Also $K > 0$ since $a_N > 0$ and $r^N > 0$. Since $(a_n)$ is a positive sequence of real numbers then $\mid a_n \mid = a_n$ and so:
(4)
\begin{align} \mid a_n \mid ≤ K r^n \end{align}
• Invoking lemma 1, we conclude that $\lim_{n \to \infty} a_n = 0$. $\blacksquare$

## Example 1

Let $0 < a < 1$. Determine if the sequence $\left ( n^2 a^n \right)$ converges.

Notice that this example pertains to a sequence of positive real numbers. Applying the ratio test we get that $\lim_{n \to \infty} \frac{(n+1)^2 a^{n+1}}{n^2 a^n} = \lim_{n \to \infty} \frac{an^2 + 2an + 1}{n^2} = \lim_{n \to \infty} a + \frac{2a}{n} + \frac{1}{n^2} = a < 1$.

By the ratio test we have that $\lim_{n \to \infty} n^2a_n = 0$.

## Example 2

Let $b > 1$. Determine if the sequence $\left ( \frac{b^n}{n!} \right)$ converges.

Notice that this example pertains to a sequence of positive real numbers. Applying the ratio test we get that $\lim_{n \to \infty} \frac{b^{n+1}}{(n+1)!} \cdot \frac{n!}{b^n} = \lim_{n \to \infty} \frac{b}{n+1} = \lim_{n \to \infty} \frac{ \frac{b}{n} }{ \frac{n}{n} + \frac{1}{n}} = \lim_{n \to \infty} \frac{ \frac{b}{n}}{1 + \frac{1}{n}}$. Consider the sequences $\left ( \frac{b}{n} \right )$ and $\left ( 1 + \frac{1}{n} \right )$ both of which converge to $0$ and $1$ respectively. Thus:

(5)
\begin{align} \lim_{n \to \infty} \frac{ \frac{b}{n}}{1 + \frac{1}{n}} = \frac{\lim_{n \to \infty} \frac{b}{n}}{\lim_{n \to \infty} 1 + \frac{1}{n}} = \frac{0}{1} = 0 \end{align}

So by the ratio test, $\left ( \frac{b^n}{n!} \right)$ converges to $0$.