The Root Test
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 5.6: “Ratio and Root Tests” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/5-6-ratio-and-root-tests |
| Textbook used in class | Stewart, Calculus, Section 11.6: “Absolute Convergence and the Ratio and Root Tests” (Example 4) |
Opening Scenario
The Ratio Test compares consecutive terms; the Root Test takes the $n$-th root of the $n$-th term. Both detect geometric-like behavior, but the Root Test is more direct when the entire term is raised to the $n$-th power -- then the $n$-th root collapses the exponent and the computation is immediate.
Quick Reference
Root Test. Let $\displaystyle L = \lim_{n\to\infty} \sqrt[n]{|a_n|} = \lim_{n\to\infty} |a_n|^{1/n}$.
- If $L < 1$: $\sum a_n$ converges absolutely.
- If $L > 1$ (or $L = \infty$): $\sum a_n$ diverges.
- If $L = 1$: inconclusive.
Best for: terms of the form $(f(n))^n$ where $f(n)$ is a rational function of $n$. Useful fact: $\displaystyle\lim_{n\to\infty} n^{1/n} = 1$ and $\displaystyle\lim_{n\to\infty} (c)^{1/n} = 1$ for any constant $c > 0$.
Key Concepts
1. When to Reach for the Root Test
If the general term has the form $a_n = \bigl(b_n\bigr)^n$, then $|a_n|^{1/n} = |b_n|$, and the limit $\lim |b_n|$ is the computation. The ratio test would require computing $a_{n+1}/a_n = (b_{n+1})^{n+1}/(b_n)^n$, which is messier.
2. Applying the Test
Example 1. Does $\displaystyle\sum_{n=1}^{\infty} \left(\frac{2n+3}{3n+2}\right)^n$ converge? (Stewart 11.6, Example 4.)
$|a_n|^{1/n} = \dfrac{2n+3}{3n+2} \to \dfrac{2}{3} < 1$.
$L = 2/3 < 1$: the series converges absolutely.
Boxed answer: Converges absolutely.
Example 2. Does $\displaystyle\sum_{n=1}^{\infty} \frac{n^n}{3^n \cdot n!}$ converge?
Here $|a_n|^{1/n} = \dfrac{n}{3 \cdot (n!)^{1/n}}$. By Stirling’s approximation, $(n!)^{1/n} \approx n/e$, so $|a_n|^{1/n} \approx \dfrac{n}{3 \cdot n/e} = \dfrac{e}{3} \approx 0.906 < 1$.
$L = e/3 < 1$: converges absolutely.
(The ratio test also works here but requires more steps.)
3. Root Test vs. Ratio Test
Both tests are inconclusive when $L = 1$. For terms involving $n^n$ or $(n/(n+1))^n$, the root test is cleaner. For terms involving $n!$ or products of polynomials and exponentials, the ratio test is usually better. When in doubt, try the one that produces a simpler computation.
confusing $n^{1/n}$ with $1/n$. $\lim_{n\to\infty} n^{1/n} = 1$, not $0$. Taking the $n$-th root of $n$ does not send it to zero; it sends it to $1$. This limit is counterintuitive and is worth computing once: $n^{1/n} = e^{(\ln n)/n} \to e^0 = 1$. Students who write $n^{1/n} \to 0$ will get wrong answers on root-test problems involving polynomial factors.
Common Errors Summary
| Error | Correction |
|---|---|
| Writing $n^{1/n} \to 0$ | $\lim n^{1/n} = 1$; use the exponential: $n^{1/n} = e^{(\ln n)/n} \to e^0 = 1$ |
| Applying the root test to $\sum 1/n^p$ and concluding convergence from $L = 1$ | $L = 1$ is always inconclusive; use $p$-series test for $\sum 1/n^p$ |
| Forgetting to take absolute value before the root | $|a_n|^{1/n}$ uses absolute value, important for alternating-like terms |
Common Misconceptions
$\lim_{n\to\infty} n^{1/n} = 0$.
This is the concept-image-conflicts-definition error about the behavior of $n^{1/n}$. Taking the $n$-th root of $n$ does not send $n$ to zero; it sends it to $1$. The proof uses the exponential: $n^{1/n} = e^{(\ln n)/n}$, and since $(\ln n)/n \to 0$ as $n \to \infty$, the limit is $e^0 = 1$. This error causes wrong root-test computations whenever a polynomial factor appears inside an $n$-th power.
$L = 1$ from the root test means the series converges.
This is the limit-equals-function-value error, exactly parallel to the same error with the ratio test. When the root test gives $L = 1$, no conclusion about convergence or divergence is possible. Any $p$-series with terms of the form $(1/n^p)^{1/n} \to 1$ illustrates this: both convergent and divergent $p$-series produce $L = 1$.
Leveled Practice
Problem 1. Use the root test on $\displaystyle\sum_{n=1}^{\infty} \left(\frac{n}{2n+1}\right)^n$.
Show answer
$|a_n|^{1/n} = \dfrac{n}{2n+1} \to \dfrac{1}{2} < 1$. Converges absolutely.
Problem 2. Use the root test on $\displaystyle\sum_{n=2}^{\infty} \left(\frac{1}{\ln n}\right)^n$.
Show answer
$|a_n|^{1/n} = \dfrac{1}{\ln n} \to 0 < 1$. Converges absolutely.