Root Test for MATH 101 A
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Understanding the Root Test
The Root Test is closely related to the Ratio Test — both compare your series to a geometric series to determine convergence. While the Ratio Test asks "how does each term compare to the previous one?", the Root Test asks a slightly different question: "what's the average decay rate per term?"
The Root Test is especially powerful when terms involve $n$th powers, like $a^n$ or expressions raised to the $n$th power.
The Conceptual Insight: What's Really Happening?
When you compute $\sqrt[n]{|a_n|}$, you're finding the geometric mean decay rate of the terms.
Think of it this way: if you had a geometric series $\sum r^n$, then:
$$\sqrt[n]{|r^n|} = |r|$$
So the $n$th root extracts the "base rate" of decay. For more complicated series, this rate might change with $n$, which is why we take the limit:
$$L = \lim_{n \to \infty} \sqrt[n]{|a_n|}$$
The key insight: If $L < 1$, then for large $n$, the terms $|a_n|$ are eventually smaller than $L^n$ for some $L < 1$. The series behaves like a convergent geometric series.
The Root Test (Statement)
For a series $\sum a_n$, compute:
$$L = \lim_{n \to \infty} \sqrt[n]{|a_n|}$$
Conclusion:
- If $L < 1$: the series converges absolutely
- If $L > 1$ (or $L = \infty$): the series diverges
- If $L = 1$: the test is inconclusive
Why Does This Work?
If $L < 1$, pick any number $r$ with $L < r < 1$.
For sufficiently large $n$, we have $\sqrt[n]{|a_n|} < r$, which means:
$$|a_n| < r^n$$
Since $\sum r^n$ converges (geometric series with $|r| < 1$), by the Comparison Test, $\sum |a_n|$ converges.
Similarly, if $L > 1$, then $|a_n| > 1$ for large $n$, so the terms don't approach zero and the series diverges.
When to Use the Root Test
The Root Test excels when you have:
- Terms of the form $(f(n))^n$ — the $n$th root simplifies beautifully
- Expressions like $\left(\frac{n}{n+1}\right)^n$ or $\left(\frac{2n+1}{3n}\right)^n$
- Anything where $n$ appears in the exponent
The Root Test struggles with:
- Factorials — Ratio Test is usually easier
- Pure polynomials like $\frac{1}{n^p}$ — always gives $L = 1$
A Useful Limit to Know
Many Root Test problems involve this classic limit:
$$\lim_{n \to \infty} \sqrt[n]{n} = 1$$
More generally:
$$\lim_{n \to \infty} \sqrt[n]{n^k} = 1 \text{ for any fixed } k$$
This tells us that polynomial factors don't affect the Root Test result — only exponential behavior matters.
Determine if $\displaystyle\sum_{n=1}^{\infty} \left(\frac{2n+1}{3n+1}\right)^n$ converges or diverges.
Step 1: Set up the root
$$\sqrt[n]{|a_n|} = \sqrt[n]{\left(\frac{2n+1}{3n+1}\right)^n} = \frac{2n+1}{3n+1}$$
Step 2: Take the limit
$$L = \lim_{n \to \infty} \frac{2n+1}{3n+1} = \lim_{n \to \infty} \frac{2 + \frac{1}{n}}{3 + \frac{1}{n}} = \frac{2}{3}$$
Step 3: Conclude
Since $L = \frac{2}{3} < 1$, the series converges absolutely.
Conceptual check: The terms behave like $\left(\frac{2}{3}\right)^n$ for large $n$ — a geometric series with ratio $\frac{2}{3}$.
Determine if $\displaystyle\sum_{n=1}^{\infty} \left(\frac{n+2}{n}\right)^n$ converges or diverges.
Step 1: Set up the root
$$\sqrt[n]{|a_n|} = \sqrt[n]{\left(\frac{n+2}{n}\right)^n} = \frac{n+2}{n} = 1 + \frac{2}{n}$$
Step 2: Take the limit
$$L = \lim_{n \to \infty} \left(1 + \frac{2}{n}\right) = 1$$
Wait — this gives $L = 1$, which is inconclusive!
Step 3: Try another approach
Actually, let's think more carefully. We know that:
$$\lim_{n \to \infty} \left(1 + \frac{2}{n}\right)^n = e^2 \approx 7.39$$
So the terms $a_n \to e^2 \neq 0$. By the Divergence Test, the series diverges.
Determine if $\displaystyle\sum_{n=1}^{\infty} \frac{n^3}{2^n}$ converges or diverges.
Step 1: Set up the root
$$\sqrt[n]{|a_n|} = \sqrt[n]{\frac{n^3}{2^n}} = \frac{\sqrt[n]{n^3}}{\sqrt[n]{2^n}} = \frac{\sqrt[n]{n^3}}{2}$$
Step 2: Take the limit
Using $\lim_{n \to \infty} \sqrt[n]{n^3} = 1$:
$$L = \lim_{n \to \infty} \frac{\sqrt[n]{n^3}}{2} = \frac{1}{2}$$
Step 3: Conclude
Since $L = \frac{1}{2} < 1$, the series converges absolutely.
Conceptual insight: The polynomial $n^3$ doesn't affect the Root Test — only the $2^n$ in the denominator matters.
Determine if $\displaystyle\sum_{n=1}^{\infty} \left(\frac{1}{n}\right)^n$ converges or diverges.
Step 1: Set up the root
$$\sqrt[n]{|a_n|} = \sqrt[n]{\left(\frac{1}{n}\right)^n} = \frac{1}{n}$$
Step 2: Take the limit
$$L = \lim_{n \to \infty} \frac{1}{n} = 0$$
Step 3: Conclude
Since $L = 0 < 1$, the series converges absolutely.
Conceptual insight: When $L = 0$, the series converges very fast — terms shrink faster than any geometric series.
Determine if $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges using the Root Test.
Step 1: Set up the root
$$\sqrt[n]{|a_n|} = \sqrt[n]{\frac{1}{n^2}} = \frac{1}{\sqrt[n]{n^2}} = \frac{1}{(\sqrt[n]{n})^2}$$
Step 2: Take the limit
Using $\lim_{n \to \infty} \sqrt[n]{n} = 1$:
$$L = \lim_{n \to \infty} \frac{1}{(\sqrt[n]{n})^2} = \frac{1}{1^2} = 1$$
Step 3: Conclude
The Root Test is inconclusive ($L = 1$).
We know this series converges (p-series with $p = 2 > 1$), but the Root Test can't detect it. Use the p-test instead!
Determine if $\displaystyle\sum_{n=1}^{\infty} \frac{n!}{n^n}$ converges or diverges.
Using the Root Test:
$$\sqrt[n]{|a_n|} = \sqrt[n]{\frac{n!}{n^n}} = \frac{\sqrt[n]{n!}}{n}$$
This requires knowing $\sqrt[n]{n!} \approx \frac{n}{e}$ (from Stirling's approximation):
$$L = \lim_{n \to \infty} \frac{n/e}{n} = \frac{1}{e} < 1$$
Conclusion: The series converges absolutely.
Note: For factorials, the Ratio Test is often simpler:
$$\frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!} = \frac{n^n}{(n+1)^n} = \left(\frac{n}{n+1}\right)^n \to \frac{1}{e}$$
Both tests give $L = \frac{1}{e}$, confirming convergence.
Root Test vs Ratio Test: When to Use Which?
| Series Type | Better Test | Why |
|---|---|---|
| $(f(n))^n$ | Root Test | $n$th root cancels the $n$th power |
| Factorials $n!$ | Ratio Test | Factorial ratio simplifies to $(n+1)$ |
| Mixed $n! \cdot r^n$ | Ratio Test | Clean cancellation |
| $\left(\frac{an+b}{cn+d}\right)^n$ | Root Test | Direct simplification |
General rule: If you see $n$ as an exponent, try Root Test first. If you see factorials, try Ratio Test first.
Summary: The Mental Model
Think of $L = \lim_{n \to \infty} \sqrt[n]{|a_n|}$ as extracting the "effective base" of the series:
| Value of $L$ | What it means | Conclusion |
|---|---|---|
| $L = 0$ | Terms shrink faster than any geometric | Converges (fast!) |
| $0 < L < 1$ | Behaves like geometric with ratio $L$ | Converges |
| $L = 1$ | Boundary case — can't tell | Inconclusive |
| $L > 1$ | Behaves like growing geometric | Diverges |
Common Mistakes and Misunderstandings
❌ Mistake: Forgetting to take the $n$th root
Wrong: Computing $|a_n|$ instead of $\sqrt[n]{|a_n|}$.
Why it's wrong: The whole point of the Root Test is extracting the $n$th root to find the decay rate.
Correct: Always compute $\sqrt[n]{|a_n|}$ first, then take the limit.
❌ Mistake: Using Root Test on pure polynomials
Wrong approach: Trying the Root Test on $\sum \frac{1}{n^3}$.
Why it's wrong: Polynomial series always give $L = 1$, making the test useless for these.
Correct: Use the p-test for $\sum \frac{1}{n^p}$. Save the Root Test for exponential-type series.
❌ Mistake: Confusing $\sqrt[n]{n} \to 1$ with $\sqrt[n]{n^n} \to n$
Wrong: Thinking $\sqrt[n]{n^n} = 1$.
Why it's wrong: $\sqrt[n]{n^n} = n$, not $1$. The exponent must match $n$ to cancel.
Correct: $\sqrt[n]{n^n} = n$, $\sqrt[n]{n^k} \to 1$ for fixed $k$, and $\sqrt[n]{c^n} = c$ for constant $c$.
❌ 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 Root Test gives NO information.
Correct: When $L = 1$, state "Root Test inconclusive" and try another test (p-test, comparison, etc.).
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