# The Ratio Test for Positive Series

We will now look at yet another test to determine whether a series is convergent or divergent known as the **Ratio Test**.

Theorem 1 (The Ratio Test for Positive Series): Suppose that $\{ a_n \}$ is an ultimately positive sequence and that $\lim_{n \to \infty} \frac{a_{n+1}}{a_{n}} = \rho$ where $\rho$ is a real number or $\rho$ is $\infty$. If $0 ≤ \rho < 1$ then the series $\sum_{n=1}^{\infty} a_n$ is convergent. If $1 < \rho ≤ \infty$ then the series $\sum_{n=1}^{\infty} a_n$ is divergent. If $\rho = 1$ then the ratio test fails as it provides no useful information regarding the convergence or divergence of a series. |

**Proof of Theorem:**Suppose that $0 ≤ \rho < 1$ and choose some $r \in \mathbb{R}$ such that $\rho < r < 1$. We note that $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \rho$ and so therefore we have that $\frac{a_{n+1}}{a_n} ≤ r$ or rather $a_{n+1} ≤ ra_n$ for $n ≥ N$ for some $N \in \mathbb{N}$ (i.e, for $n$ sufficiently large). Furthermore $a_{N+1} ≤ ra_N$, $a_{N+2} ≤ ra_{N+1} ≤ r^2 a_N$, …, $a_{N+k} ≤ r^k a_n$ for $k = 1, 2, ...$.

- Therefore the series $\sum_{n=N}^{\infty} a_n$ converges by comparison with the convergent geometric series since $0 ≤ \rho < r < 1$. Of course this implies that $\sum_{n=1}^{\infty} a_n$ is also convergent.

- Now let's suppose that $\rho > 1$ and pick some $r \in \mathbb{R}$ such that $1 < r < \rho$. Now since $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \rho$ it follows that $\frac{a_{n+1}}{a_n} ≥ r$ or rather $a_{n+1} ≥ ra_n$ if $n ≥ N$ for some $n \in \mathbb{N}$. Using the same argument as before we note that $a_{N+k} ≥ r^ka_N$. Since $1 < r < \rho$ it follows by the comparison theorem with geometric series that $\sum_{n=N}^{\infty} a_n$ is divergent and similarly $\sum_{n=1}^{\infty} a_n$ is also divergent. $\blacksquare$

## Example 1

**Using the ratio test determine whether the series $\sum_{n=1}^{\infty} \frac{1}{n!}$ is convergent or divergent.**

Using the ratio test, we will want to determine $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{\frac{1}{(n+1)!}}{\frac{1}{n!}}$.

(1)So $\rho = 0$ and $0 ≤ \rho < 1$ so by the ratio test, the series $\sum_{n=1}^{\infty} \frac{1}{n!}$ is convergent.

## Example 2

**Using the ratio test determine whether the series $\sum_{n=1}^{\infty} \frac{n}{n!}$ is convergent or divergent.**

Using the ratio test we get that:

(2)So therefore since $\rho = 0$ and $0 ≤ \rho < 1$, by the ratio test the series $\sum_{n=1}^{\infty} \frac{n}{n!}$ is convergent.

## Example 3

**Using the ratio test determine whether the series $\sum_{n=1}^{\infty} \frac{2n! + (n+1)!}{en!}$ is convergent or divergent.**

Notice that if we apply the ratio test then:

(3)So the ratio test yields no information. However, we can simplify $\frac{2n! + (n+1)!}{en!}$ as $\frac{2n! + (n+1)!}{en!} = \frac{2}{e} + n$. Notice that $\lim_{n \to \infty} \frac{2}{e} + n = \infty$. By the divergence test, we see that $\sum_{n=1}^{\infty} \frac{2n! + (n+1)!}{en!}$ diverges.

*It is important to note that while the ratio test is extremely useful - there are times when it is unnecessary to apply it. Examine alternative options to determining convergence/divergence beforehand.*