Why nth roots reveal convergence
A series can be difficult to classify by inspecting its terms directly. The root test addresses this by measuring the exponential scale of the terms: if their nth roots eventually stay below a number less than one, the terms are bounded by a convergent geometric sequence. If the roots eventually stay above one, the terms cannot approach zero.
The absolute value is essential. A series may have negative or complex terms, but the root test compares the nonnegative quantities \(|a_n|\). The test therefore first determines absolute convergence, not whether signs or complex directions create cancellation.
The value \(L\) is nonnegative, and it may be \(+\infty\). The conclusion for \(L>1\) applies to that case as well. The test does not say that every series with \(L=1\) diverges or that every such series converges: it is simply inconclusive there.
The geometric comparison behind the test
The key fact is that a limit gives an eventual bound. When \(L<1\), choose a real number \(q\) with \(L<q<1\). Eventually, \(|a_n|^{1/n}<q\), so raising both nonnegative sides to the positive integer power \(n\) gives \(|a_n|<q^n\). The geometric series \(\sum q^n\) converges because \(0<q<1\). Thus the tail of \(\sum |a_n|\) is bounded term by term by a convergent geometric series.
For \(L>1\), choose \(c\) with \(1<c<L\) if \(L\) is finite. If \(L=+\infty\), choose, for example, \(c=2\). Eventually, \(|a_n|^{1/n}>c\), and hence \(|a_n|>c^n\). In particular, the terms do not approach zero. This conclusion concerns their absolute values; it does not assert that \(a_n>c^n\), which would be false for negative terms and would not be meaningful as an order comparison for complex terms.
Proof of the root test
Here is the complete argument. It uses the comparison test for nonnegative series and the geometric-series formula: \(\sum_{n=1}^{\infty}q^n=q/(1-q)\) when \(0<q<1\). Absolute convergence of \(\sum a_n\) means convergence of \(\sum |a_n|\).
Choose \(q\) such that \(L<q<1\). By the definition of the limit, there is an index \(N\) such that for every \(n\geq N\), \(\bigl||a_n|^{1/n}-L\bigr|<q-L\). Therefore \(|a_n|^{1/n}<q\), and raising to the power \(n\) yields \(|a_n|<q^n\).
For \(n\geq N\), \(0\leq |a_n|\leq q^n\), while \(\sum_{n=N}^{\infty}q^n=q^N/(1-q)\) is finite. The comparison test gives convergence of \(\sum_{n=N}^{\infty}|a_n|\). Adding the finite initial sum \(\sum_{n=1}^{N-1}|a_n|\) preserves convergence, so \(\sum |a_n|\) converges.
Choose \(c\) with \(1<c<L\) when \(L\) is finite; if \(L=+\infty\), take \(c=2\). The limit definition implies that eventually \(|a_n|^{1/n}>c\). Thus \(|a_n|>c^n\). Since \(c>1\), \(c^n\geq c>0\) for all positive integers \(n\), so \(|a_n|\) cannot tend to zero.
A necessary condition for convergence of \(\sum a_n\) is \(a_n\to0\), which also requires \(|a_n|\to0\). The bound in Step 3 contradicts this, so the series diverges. If \(L=1\), neither argument supplies a geometric bound with ratio below one or a positive lower bound growing above one; the test alone gives no conclusion.
This proves both decisive cases. Notice that absolute convergence in the first case is stronger than ordinary convergence: it ensures convergence regardless of the signs or complex arguments of the terms.
Worked examples
Worked Example: A polynomial factor over an exponential
Classify \(\sum_{n=1}^{\infty}\frac{n^2}{3^n}\) using the root test.
Set \(a_n=n^2/3^n\), which is positive. Its nth root is
To evaluate \(n^{2/n}\), take logarithms: \(\ln(n^{2/n})=2\ln(n)/n\), which tends to zero; therefore \(n^{2/n}\to e^0=1\). It follows that
The root test proves absolute convergence. As a check on the resulting type of bound, choose \(q=1/2\), which satisfies \(1/3<1/2<1\). Eventually \(|a_n|^{1/n}<1/2\), so \(|a_n|<(1/2)^n\); the comparison series has tail \(\sum_{n=N}^{\infty}(1/2)^n=(1/2)^N/(1-1/2)=2^{1-N}\), which is finite. This confirms the geometric-comparison mechanism used in the proof.
Worked Example: Alternating terms with magnitude growing exponentially
Determine whether \(\sum_{n=1}^{\infty}(-1)^n(3/2)^n\) converges.
Here \(a_n=(-1)^n(3/2)^n\), so \(|a_n|=(3/2)^n\). Therefore
The root test says the series diverges. The term condition verifies this directly: \(|a_n|=(3/2)^n\geq3/2\) for every \(n\geq1\), so \(|a_n|\not\to0\). In fact, the signs alternate but the term magnitudes grow; alternation cannot make a series converge when its terms fail to approach zero.
Worked Example: The boundary value does not decide the answer
Compare \(\sum_{n=1}^{\infty}1/n\) and \(\sum_{n=1}^{\infty}1/n^2\) with the root test.
For the first series, \(a_n=1/n\), so
The root test returns \(L=1\), and therefore gives no conclusion. This series diverges: for each integer \(k\geq1\), the block with \(2^{k-1}<n\leq2^k\) contains \(2^{k-1}\) terms, each at least \(1/2^k\), so that block sums to at least \(1/2\). The partial sums therefore grow without bound.
For the second series, \(a_n=1/n^2\), and
Again \(L=1\), so the root test gives no conclusion. This series converges: for \(n\geq2\), \(1/n^2\leq1/[n(n-1)]=1/(n-1)-1/n\), and hence for every \(M\geq2\), \(\sum_{n=2}^{M}1/n^2\leq1-1/M<1\). Its partial sums are increasing and bounded, so they converge. The two checks show why the boundary case is genuinely inconclusive: the same root limit occurs for a convergent series and a divergent one.
Conditions and common pitfalls
The root test is most useful when taking nth roots simplifies the expression for the terms. Its conclusion depends on the limit existing as a finite number or \(+\infty\). The later version using limit superior handles cases where the ordinary limit does not exist; that is a different criterion and is not needed for the proof here.
| Root limit | What follows |
|---|---|
| \(L<1\) | \(\sum |a_n|\) converges, so \(\sum a_n\) converges absolutely. |
| \(L>1\) | \(|a_n|\) does not tend to zero, so \(\sum a_n\) diverges. |
| \(L=1\) | No conclusion from the root test; other information is required. |
A common error in the \(L>1\) case is to write \(a_n>c^n\). The root bound gives \(|a_n|>c^n\), not an inequality for \(a_n\) itself. For example, with \(a_n=(-1)^n2^n\), the false claim \(a_n>1\) fails for every odd \(n\), even though \(|a_n|=2^n>1\) and the series diverges. The correction is to use the absolute values throughout: \(|a_n|^{1/n}>c>1\) implies \(|a_n|>c^n\), and thus the terms do not tend to zero.
A second pitfall is interpreting \(L=1\) as a borderline verdict of divergence. The two series in the boundary example have the same root limit but different outcomes. A root limit of one means only that this test cannot distinguish them; it does not rule out a conclusion from another appropriate method.
Why the root test matters
The proof explains the test's strength: nth roots convert complicated term formulas into a comparison of exponential growth rates. A limit below one eventually places the terms beneath a decaying geometric sequence. A limit above one forces their magnitudes to stay away from zero and grow. This is a compact way to recognize when exponential behavior settles a convergence question without needing to sum the series.
Check Your Understanding
Use the root-test proof and its conditions to answer these questions.
- In the case \(L<1\), why must the comparison number \(q\) be chosen with \(L<q<1\)?
- If \(|a_n|^{1/n}>c>1\) eventually, what inequality follows for \(|a_n|\), and why does it imply divergence?
- For \(a_n=(-1)^n(1/4)^n\), find the root limit and state the test's conclusion.
- What does the root test say about \(\sum 1/n^3\), and why does that not establish convergence or divergence?
- Why is it incorrect to infer \(a_n>c^n\) for a series whose terms may be negative or complex?
