The Ratio Test for Convergence

Core Concepts

A powerful test, especially for series with factorials or exponential terms. It determines convergence by analyzing the limit of the ratio of consecutive terms.

The Ratio Test for Convergence

1 of 9

State the formula used in the Ratio Test for a series $\sum a_n$.

You must compute the limit $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$.

In the Ratio Test, if $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1$, what is the conclusion?

The series $\sum a_n$ is absolutely convergent (and therefore converges).

In the Ratio Test, if $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| > 1$ or $L = \infty$, what is the conclusion?

The series $\sum a_n$ diverges.

In the Ratio Test, if $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 1$, what is the conclusion?

The test is inconclusive. The series may converge or diverge. You must use a different test.

The Ratio Test is particularly effective for series containing which two types of mathematical expressions?

It is most effective for series involving factorials (like $n!$) and exponentials (like $c^n$).

Apply the Ratio Test to $\sum_{n=1}^{\infty} \frac{3^n}{n!}$.

Compute $L = \lim_{n \to \infty} \left| \frac{3^{n+1}/(n+1)!}{3^n/n!} \right| = \lim_{n \to \infty} \frac{3^{n+1}}{3^n} \cdot \frac{n!}{(n+1)!} = \lim_{n \to \infty} 3 \cdot \frac{1}{n+1} = 0$. Since $L = 0 < 1$, the series converges.

Apply the Ratio Test to $\sum_{n=1}^{\infty} \frac{n^2}{2^n}$.

Compute $$L = \lim_{n \to \infty} \left| \frac{(n+1)^2/2^{n+1}}{n^2/2^n} \right| = \lim_{n \to \infty} \frac{(n+1)^2}{n^2} \cdot \frac{2^n}{2^{n+1}} = \lim_{n \to \infty} \left(1+\frac{1}{n}\right)^2 \cdot \frac{1}{2} = 1^2 \cdot \frac{1}{2} = \frac{1}{2}$$ Since $L = 1/2 < 1$, the series converges.

Why is the Ratio Test inconclusive for any p-series, $\sum \frac{1}{n^p}$?

For a p-series, the limit of the ratio is always $1$. $$\lim_{n \to \infty} \frac{1/(n+1)^p}{1/n^p} = \lim_{n \to \infty} (\frac{n}{n+1})^p = 1^p = 1$$ Since the limit is $1$, the test provides no information.

What does $\frac{(n+1)!}{n!}$ simplify to?

It simplifies to $n+1$, since $$(n+1)! = (n+1) \cdot n \cdot (n-1) \cdots 1 = (n+1) \cdot n!$$

The Ratio Test for Convergence

Confirm Action

Prompt