Ratio Test for Master Mathematics

Understanding the Ratio Test

The Ratio Test is one of the most powerful tools for determining convergence. It works especially well for series involving factorials, exponentials, and powers — situations where comparison tests become awkward. But beyond just being a "test to memorize," the Ratio Test reveals something deep about how fast terms are shrinking.

The Conceptual Insight: What's Really Happening?

Before we state the formula, let's understand what we're actually measuring.

When you compute $\displaystyle\frac{a_{n+1}}{a_n}$, you're asking: "How much smaller (or larger) is each term compared to the previous one?"

• If this ratio is consistently around $0.5$, each term is about half the previous term
• If this ratio is consistently around $0.9$, each term is about 90% of the previous term
• If this ratio is consistently around $1.2$, each term is about 20% bigger than the previous term

The key insight: If terms eventually shrink by at least some fixed percentage each step (ratio < 1), the series converges. If terms eventually grow by any fixed percentage each step (ratio > 1), the series diverges.

This is just like geometric series! If the common ratio has $|r| < 1$, the geometric series converges. The Ratio Test extends this idea to series that aren't exactly geometric but behave like geometric series for large $n$.

Why Taking the Limit Matters

For most series, the ratio $\frac{a_{n+1}}{a_n}$ isn't constant — it changes with $n$. That's why we take the limit:

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

This limit $L$ tells us: "In the long run, what's the shrinking/growing factor?"

If $L = 0.7$, then for very large $n$, each term is roughly $70\%$ of the previous term. The tail of the series behaves like a geometric series with ratio $0.7$, which converges.

If $L = 1.3$, then for very large $n$, each term is roughly $130\%$ of the previous term. The terms are growing, so the series must diverge.

The Ratio Test (Statement)

For a series $\sum a_n$ with $a_n \neq 0$, compute:

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

Conclusion:

• If $L < 1$: the series converges absolutely
• If $L > 1$ (or $L = \infty$): the series diverges
• If $L = 1$: the test is inconclusive (could go either way)

Why $L = 1$ Is Inconclusive

When $L = 1$, the terms are shrinking (or staying the same) but barely. The series is right on the boundary between convergence and divergence.

Example: Both $\sum \frac{1}{n}$ (diverges) and $\sum \frac{1}{n^2}$ (converges) give $L = 1$ in the Ratio Test.

For $\sum \frac{1}{n}$: $$\frac{a_{n+1}}{a_n} = \frac{1/(n+1)}{1/n} = \frac{n}{n+1} \to 1$$

For $\sum \frac{1}{n^2}$: $$\frac{a_{n+1}}{a_n} = \frac{1/(n+1)^2}{1/n^2} = \frac{n^2}{(n+1)^2} \to 1$$

Same limit, opposite conclusions! When $L = 1$, you need a different test.

When to Use the Ratio Test

The Ratio Test excels when you have:

• Factorials ($n!$) — the $(n+1)!/n! = n+1$ cancellation is clean
• Exponentials ($a^n$) — the $a^{n+1}/a^n = a$ cancellation is clean
• Powers of $n$ combined with exponentials ($n^k \cdot r^n$)

The Ratio Test struggles with:

• Pure powers of $n$ like $\frac{1}{n^p}$ — always gives $L = 1$
• Algebraic expressions without exponentials or factorials

The Ratio Test and Absolute Convergence

Notice that the Ratio Test always uses absolute values:

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

This means when the Ratio Test says "converges," it means converges absolutely. You get absolute convergence for free!

Summary: The Mental Model

Think of the limit $L$ as the "long-run shrinking factor":

Common Mistakes and Misunderstandings

❌ Mistake: Using Ratio Test on pure polynomials

Wrong approach: Trying Ratio Test on $\sum \frac{1}{n^3}$

Why it's wrong: Pure polynomial terms always give $L = 1$. The Ratio Test will never work on these.

Correct: Use the p-test: $p = 3 > 1$, so it converges.

❌ Mistake: Forgetting absolute values

Wrong: Computing $\frac{a_{n+1}}{a_n}$ without absolute values for alternating series.

Why it's wrong: For $\sum \frac{(-1)^n}{n!}$, the ratio alternates in sign. You need $\left|\frac{a_{n+1}}{a_n}\right|$.

Correct: Always use $\left|\frac{a_{n+1}}{a_n}\right|$ in the Ratio Test.

❌ Mistake: Concluding from $L = 1$

Wrong: "$L = 1$, so the series converges" or "$L = 1$, so it diverges."

Why it's wrong: $L = 1$ means the test gives NO information. You must use another test.

Correct: When $L = 1$, state "Ratio Test inconclusive" and try p-test, comparison, or another method.