Why use a limit superior?
A standard version of the root test is stated when the sequence of nth roots of the term magnitudes has a limit. But those roots need not settle toward a single value: they can oscillate between different values. The limit superior captures the eventual upper level of such a sequence and gives a root test that still applies.
For a series \(\sum_{n=1}^{\infty} a_n\), the root test studies \(\sqrt[n]{|a_n|}\). The absolute value matters: the test gives a conclusion about absolute convergence, even if the original terms have alternating signs or are complex.
In this tutorial, set \(x_n=\sqrt[n]{|a_n|}\) and write \(L=\limsup_{n\to\infty}\sqrt[n]{|a_n|}\). If \(a_n=0\), its nth root is \(0\). The value \(L\) is the eventual upper exponential scale of the term magnitudes; unlike an ordinary limit, it exists even when the roots oscillate.
The root test with limit superior
“Gives no conclusion” means that the test alone does not determine convergence or divergence. It does not mean that the series is divergent, and it does not mean that the series converges. Both outcomes are possible when \(L=1\).
Why the test works
The proof uses two facts about the tail suprema. If \(L<r\), then eventually every nth root is less than \(r\). If \(L>r\), then infinitely many nth roots exceed \(r\). Choosing \(r\) on the appropriate side of \(1\) turns these facts into comparison with geometric terms or failure of the term test.
Choose a real number \(r\) with \(L<r<1\). By the definition of limit superior, for all sufficiently large \(n\), \(\sqrt[n]{|a_n|}<r\). Raising both sides to the positive integer power \(n\) gives \(|a_n|<r^n\).
The geometric series \(\sum r^n\) converges because \(0<r<1\). For all sufficiently large \(n\), \(0\le |a_n|<r^n\), so the comparison test gives \(\sum |a_n|<\infty\). Thus \(\sum a_n\) converges absolutely.
Choose \(r\) with \(1<r<L\). Infinitely many indices satisfy \(\sqrt[n]{|a_n|}>r\); otherwise all sufficiently late roots would be at most \(r\), forcing \(L\le r\). At each such index, \(|a_n|>r^n>1\). Therefore \(a_n\) does not tend to \(0\).
If a series converges, its terms must tend to zero. Since the terms do not tend to zero when \(L>1\), the series diverges. When \(L=1\), neither comparison above is available: there is no \(r<1\) greater than \(L\), and no \(r>1\) less than \(L\).
This proves all three cases. Notice that \(L>1\) need not mean every term is large. The proof needs only infinitely many terms that fail to approach zero.
Worked Example: Oscillating roots with limit superior below one
Worked Example: Oscillating roots with limit superior below one
Define positive terms by $$ a_n= \begin{cases} \left(\frac{3}{4}\right)^n,&n\text{ even},\\[2pt] \left(\frac{1}{2}\right)^n,&n\text{ odd}. \end{cases} $$ For even \(n\), \(\sqrt[n]{a_n}=\frac34\). For odd \(n\), \(\sqrt[n]{a_n}=\frac12\). The roots do not converge, but every tail contains even indices and odd indices, so its supremum is \(\frac34\). Hence $$ L=\limsup_{n\to\infty}\sqrt[n]{a_n}=\frac34<1. $$ The root test therefore gives absolute convergence. Directly, \(a_n\le(3/4)^n\) for every \(n\), and \(\sum_{n=1}^{\infty}(3/4)^n\) converges. This comparison confirms the conclusion.
Worked Example: A limit superior greater than one
Worked Example: A limit superior greater than one
Let $$ a_n= \begin{cases} 2^n,&n=k^2\text{ for some }k\in\mathbb{N},\\[2pt] \left(\frac12\right)^n,&\text{otherwise}. \end{cases} $$ When \(n\) is a square, \(\sqrt[n]{|a_n|}=2\); otherwise it is \(\frac12\). There are infinitely many square indices, so every tail has supremum \(2\). Thus \(L=2>1\), and the root test predicts divergence. The terms also fail the necessary term condition: along \(n=k^2\), $$ a_{k^2}=2^{k^2}\ge 2, $$ so this subsequence does not tend to zero. Therefore the series diverges.
Worked Example: Limit superior one with a convergent series
Worked Example: Limit superior one with a convergent series
Consider \(a_n=2^{-\sqrt n}\). Its nth root is $$ \sqrt[n]{a_n}=2^{-\sqrt n/n}=2^{-1/\sqrt n}\longrightarrow 1, $$ so \(L=1\). The root test is inconclusive. To determine convergence by comparison, group the indices into blocks \(k^2\le n<(k+1)^2\), for \(k\ge1\). Each block contains \((k+1)^2-k^2=2k+1\) terms. For each index in that block, \(\sqrt n\ge k\), and hence \(2^{-\sqrt n}\le2^{-k}\). Therefore the sum over the block is at most \((2k+1)2^{-k}\).
The series of these block bounds converges: for \(b_k=(2k+1)2^{-k}\), $$ \frac{b_{k+1}}{b_k} =\frac{2k+3}{2k+1}\cdot\frac12 =\frac{2k+3}{4k+2} \longrightarrow\frac12<1. $$ By the ratio test, \(\sum_{k=1}^{\infty}b_k\) converges. The block comparison proves that \(\sum_{n=1}^{\infty}2^{-\sqrt n}\) converges. This is a convergent series for which \(L=1\).
Worked Example: Limit superior one with a divergent series
Worked Example: Limit superior one with a divergent series
For \(a_n=1/n\), $$ \sqrt[n]{a_n}=n^{-1/n} =\exp\left(-\frac{\ln n}{n}\right)\longrightarrow1, $$ because \(\frac{\ln n}{n}\to0\). Thus \(L=1\), and the root test is inconclusive. The series nevertheless diverges. In each block \(2^k\le n<2^{k+1}\), there are \(2^k\) terms, and every one satisfies \(1/n\ge1/2^{k+1}\). The block sum is therefore at least $$ 2^k\cdot\frac{1}{2^{k+1}}=\frac12. $$ There are infinitely many disjoint blocks, each contributing at least \(\frac12\), so the partial sums are unbounded. Hence \(\sum_{n=1}^{\infty}1/n\) diverges. This gives the other possible outcome when \(L=1\).
A common mistake: treating one as a verdict
A frequent error is to conclude that a series diverges whenever the root-test value is \(1\). For example, the incorrect calculation might say: “For \(a_n=1/n^2\), \(\sqrt[n]{a_n}=n^{-2/n}\to1\), so the series diverges.” The root calculation is correct, but the conclusion is not: \(L=1\) is the inconclusive case, not the divergence case. In fact, \(\sum 1/n^2\) converges, while the harmonic series diverges, and both have root-test value \(1\).
Another error is to require the nth roots to have an ordinary limit before using the root test. The first worked example has no such limit, but its limit superior is well-defined and less than one. The limsup formulation is designed precisely to handle this situation.
Check Your Understanding
For each question, identify what the root test establishes and distinguish the test’s conclusion from any additional reasoning required.
- If \(\limsup\sqrt[n]{|a_n|}=0.6\), what can you conclude about \(\sum a_n\)?
- If the nth roots alternate between \(0.4\) and \(1.2\), what is their limit superior, and what does the root test say?
- Why does \(L>1\) imply that the terms fail to tend to zero?
- If \(L=1\), is the series necessarily divergent? Give the two contrasting examples from this tutorial.
- For \(a_n=3^n\) when \(n\) is a square and \(a_n=1/3^n\) otherwise, find \(L\) and state the test’s conclusion.