Math101learn.math101.caRoot Test
A rigorous, example-driven guide to root test, including hypotheses, method choice, verification, and practice.
The central idea
For $\sum a_n$, let $L=\limsup |a_n|^{1/n}$. If $L<1$, the series converges absolutely; if $L>1$, it diverges; if $L=1$, the test is inconclusive. The root test is natural when the full term is raised to the $n$th power.
Definitions, hypotheses, and notation
When the ordinary root limit exists, it equals the limsup and the familiar calculation is enough. The limsup version handles oscillating effective bases. As with the ratio test, $L<1$ yields an eventual geometric bound and therefore absolute convergence.
The root test often simplifies expressions containing $c_n^n$, while factorials usually favor ratios. It can also find power-series radii by producing a factor $|x-a|$. If a remaining root such as $n^{1/n}$ appears, use its limit one rather than treating it as exactly one.
Conceptual meaning
The nth root extracts the effective geometric factor from a term. Polynomial and constant prefactors have nth roots tending to one, leaving the exponential-scale rate that governs the tail.
A dependable method and decision rule
- Take the absolute value of the general term.
- Apply the nth root to every factor.
- Simplify powers before taking the limit or limsup.
- Compare the resulting $L$ with one.
- If $L=1$, move to comparison, integral, or another structure-appropriate test.
Fully worked example
Graphical or geometric meaning
Taking nth roots removes the steep exponential compression and reveals the underlying base. If that base settles below one, original term magnitudes fall like powers of a shrinking factor.
Common mistakes and why they fail
Verification and reasonableness checks
- Raise the simplified root expression back to the nth power.
- Compare with a geometric series whose ratio is slightly larger than $L$.
- Check that $L>1$ also makes terms fail to approach zero.
Use roots for terms raised to the index
The root test uses $L=\limsup\sqrt[n]{|a_n|}$ and is effective when most of a term is raised to the $n$th power. It gives absolute convergence for $L<1$, divergence for $L>1$, and no conclusion for $L=1$. The limsup formulation accommodates mild oscillations even when an ordinary limit fails. Constants whose $n$th roots tend to one do not affect the outcome. With a power series, retain $|x-c|$ while simplifying, solve the resulting strict inequality, and examine each boundary point by another method. A bounded sequence of roots is not enough; its limiting upper behavior must be compared specifically with one.
Before taking roots, isolate all factors that depend exponentially on $n$. Polynomial factors usually contribute an $n$th-root limit of one, which explains why the test detects exponential scale rather than lower-order growth.
Practice
- Test $\sum(3/4)^n$ by roots.
- Test $\sum[(n+1)/n]^n$.
- What does the root test say for $1/n^2$?
Answers and brief solutions
- Converges; $L=3/4$.
- It diverges because its terms tend to $e$, not zero; the root test itself is inconclusive.
- Inconclusive; $L=1$.
Connections and next steps
Try it yourself
Hints are part of learning. Open one whenever it makes the next step feel possible.
What does the root test conclude for Σ[(2n+1)/(4n)]^n?
- The nth root is (2n+1)/(4n).
- Its limit is 1/2.
- Since 1/2<1, the series converges absolutely.
End of lesson
Nice work making it this far.
Understanding grows through return visits. Save this lesson, try the practice, or continue when you are ready.
