Topic 10.8 Notes – Ratio Test for Convergence

The Ratio Test

For a series $\sum a_n$, define

$$L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|.$$

Here’s what the value of $L$ tells you:

• If $L < 1$ → the series converges absolutely
• If $L > 1$ (or $L = \infty$) → the series diverges
• If $L = 1$ → the test is inconclusive

Why this works: you’re checking how each term compares to the one before it.
If the terms are shrinking by a consistent factor less than 1, the series behaves like a geometric series with $|r| < 1$, which converges.

The diagram below summarizes the full decision process.

Ratio Test decision flowchart

How to Apply the Ratio Test

When you actually do one of these, the algebra matters.

1. Write $a_n$ clearly.
   For example, if the series is $\sum \frac{3^n}{n!}$, then
   $a_n = \frac{3^n}{n!}$.

2. Find $a_{n+1}$ by replacing every $n$ with $n+1$.
   Be careful inside exponents and factorials.

3. Form the ratio
   $$\frac{a_{n+1}}{a_n}$$

4. Simplify completely before taking the limit.
   This is where most of the magic happens:

   • $(n+1)! = (n+1)n!$
   • $c^{n+1} = c \cdot c^n$

   For the example above:

   $$\frac{a_{n+1}}{a_n} = \frac{\frac{3^{n+1}}{(n+1)!}}{\frac{3^n}{n!}} = \frac{3^{n+1}}{(n+1)!} \cdot \frac{n!}{3^n} = \frac{3}{n+1}.$$

5. Take the limit as $n \to \infty$
   $$\lim_{n\to\infty} \frac{3}{n+1} = 0.$$

Since $0 < 1$, the series converges absolutely.

Notice how everything canceled cleanly. That’s a sign you picked the right test.

When the Ratio Test Is the Best Choice

You should think “Ratio Test” almost automatically when you see:

• Factorials like $n!$
• Exponentials like $5^n$, $(-2)^n$
• Expressions raised to the $n$th power, such as $(\frac{2n}{3n+1})^n$
• Combinations of factorials and exponentials

The reason is growth rates:

$$\text{Factorials grow faster than exponentials, which grow faster than polynomials.}$$

That growth behavior is exactly what the ratio compares.

For example, if $a_n = \frac{n!}{4^n}$, the ratio becomes

$$\frac{a_{n+1}}{a_n} = \frac{(n+1)!}{4^{n+1}} \cdot \frac{4^n}{n!} = \frac{n+1}{4}.$$

As $n \to \infty$, this goes to infinity. Since $L > 1$, the series diverges.

You can see the factorial in the numerator eventually “wins.”

What If L = 1

If
$$\lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| = 1,$$ the Ratio Test gives you no information.

This often happens with:

• Rational expressions like $\frac{n+2}{n^2+1}$
• p-series forms
• Polynomial over polynomial terms

In those cases, switch to another test from Unit 10 such as comparison, limit comparison, integral test, alternating series test, or the nth-term test for divergence.

On an FRQ, it’s completely correct to write that the Ratio Test is inconclusive and then move on.

Common Mistakes

• Forgetting the absolute value.
• Plugging into the limit before simplifying.
• Making errors when finding $a_{n+1}$.
• Not canceling factorials fully.
• Saying “converges” instead of “converges absolutely.”

On multiple choice, the trap answer is often based on a tiny algebra mistake in the ratio. Slow down during simplification.