Ratio Test for MATH 101 A
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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
Determine if $\displaystyle\sum_{n=1}^{\infty} \frac{n!}{2^n}$ converges or diverges.
Step 1: Set up the ratio
$$\frac{a_{n+1}}{a_n} = \frac{(n+1)!/2^{n+1}}{n!/2^n}$$
Step 2: Simplify
$$= \frac{(n+1)!}{n!} \cdot \frac{2^n}{2^{n+1}} = (n+1) \cdot \frac{1}{2} = \frac{n+1}{2}$$
Step 3: Take the limit
$$L = \lim_{n \to \infty} \frac{n+1}{2} = \infty$$
Conclusion: Since $L = \infty > 1$, the series diverges.
Conceptual check: This makes sense! Factorials grow faster than exponentials, so the terms $\frac{n!}{2^n}$ eventually get huge.
Determine if $\displaystyle\sum_{n=1}^{\infty} \frac{n^3}{5^n}$ converges or diverges.
Step 1: Set up the ratio
$$\frac{a_{n+1}}{a_n} = \frac{(n+1)^3/5^{n+1}}{n^3/5^n}$$
Step 2: Simplify
$$= \frac{(n+1)^3}{n^3} \cdot \frac{5^n}{5^{n+1}} = \left(\frac{n+1}{n}\right)^3 \cdot \frac{1}{5}$$
Step 3: Take the limit
$$L = \lim_{n \to \infty} \left(\frac{n+1}{n}\right)^3 \cdot \frac{1}{5} = 1^3 \cdot \frac{1}{5} = \frac{1}{5}$$
Conclusion: Since $L = \frac{1}{5} < 1$, the series converges absolutely.
Conceptual insight: The ratio approaches $\frac{1}{5}$, meaning for large $n$, each term is about $\frac{1}{5}$ of the previous. The tail behaves like a geometric series with ratio $\frac{1}{5}$.
Determine if $\displaystyle\sum_{n=0}^{\infty} \frac{3^n}{n!}$ converges or diverges.
Step 1: Set up the ratio
$$\frac{a_{n+1}}{a_n} = \frac{3^{n+1}/(n+1)!}{3^n/n!}$$
Step 2: Simplify
$$= \frac{3^{n+1}}{3^n} \cdot \frac{n!}{(n+1)!} = 3 \cdot \frac{1}{n+1} = \frac{3}{n+1}$$
Step 3: Take the limit
$$L = \lim_{n \to \infty} \frac{3}{n+1} = 0$$
Conclusion: Since $L = 0 < 1$, the series converges absolutely.
Conceptual insight: The ratio $\frac{3}{n+1}$ goes to $0$, meaning eventually each term is a tiny fraction of the previous term. The factorial in the denominator dominates completely. (This series equals $e^3$!)
Determine if $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges using the Ratio Test.
Step 1: Set up the ratio
$$\frac{a_{n+1}}{a_n} = \frac{1/(n+1)^2}{1/n^2} = \frac{n^2}{(n+1)^2}$$
Step 2: Take the limit
$$L = \lim_{n \to \infty} \frac{n^2}{(n+1)^2} = \lim_{n \to \infty} \left(\frac{n}{n+1}\right)^2 = 1$$
Conclusion: The test is inconclusive.
We know this series converges (p-series with $p = 2 > 1$), but the Ratio Test can't tell us that. Use the p-test instead!
Determine if $\displaystyle\sum_{n=1}^{\infty} \frac{n^2 \cdot 3^n}{(n+1)!}$ converges or diverges.
Step 1: Set up the ratio
$$\frac{a_{n+1}}{a_n} = \frac{(n+1)^2 \cdot 3^{n+1}/(n+2)!}{n^2 \cdot 3^n/(n+1)!}$$
Step 2: Simplify piece by piece
$$= \frac{(n+1)^2}{n^2} \cdot \frac{3^{n+1}}{3^n} \cdot \frac{(n+1)!}{(n+2)!}$$
$$= \left(\frac{n+1}{n}\right)^2 \cdot 3 \cdot \frac{1}{n+2}$$
Step 3: Take the limit
$$L = \lim_{n \to \infty} \left(\frac{n+1}{n}\right)^2 \cdot \frac{3}{n+2} = 1 \cdot 0 = 0$$
Conclusion: Since $L = 0 < 1$, the series converges absolutely.
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":
| Value of $L$ | What it means | Conclusion |
|---|---|---|
| $L = 0$ | Terms shrink extremely fast | Converges (very fast!) |
| $0 < L < 1$ | Terms shrink by factor $L$ each step | Converges (like geometric) |
| $L = 1$ | Terms barely shrink | Inconclusive |
| $L > 1$ | Terms grow by factor $L$ each step | Diverges |
| $L = \infty$ | Terms explode | Diverges (fast!) |
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.
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