Measuring the Exponential Scale of a Term
The Ratio Test studies how much a term changes from one index to the next. The Root Test asks a related question: what fixed exponential scale best describes the size of the terms? For a series \(\sum_{n=1}^{\infty}a_n\), the quantity \(|a_n|^{1/n}\) measures that scale. When it stays below one, the terms are eventually bounded by a convergent geometric sequence; when it stays above one along a subsequence, the terms cannot tend to zero.
The nth-root sequence need not have a limit, so the most useful formulation uses a limit superior. For a nonnegative sequence \((x_n)\), its limit superior is the extended real number
In words, this records the largest value that the sequence continues to approach or exceed arbitrarily far out. It can be \(+\infty\). For a series, we apply this definition to \(x_n=|a_n|^{1/n}\), taking \(0^{1/n}=0\).
The Root Test does not require the terms to be eventually nonzero. If some terms are zero, their nth roots are simply zero. The theorem also includes the familiar version in which the ordinary limit \(\lim_{n\to\infty}|a_n|^{1/n}\) exists: in that case, the limit superior equals that limit.
Why Root Bounds Produce Geometric Bounds
The convergence threshold is one because powers of a number below one form a convergent geometric series. The following observation isolates the estimate that makes this comparison possible.
Proof. Write \(x_n=|a_n|^{1/n}\). Since \(\limsup x_n=\rho<q\), the definition of limit superior gives an index \(N\) such that \(\sup_{n\geq N}x_n<q\). Thus \(x_n<q\) for every \(n\geq N\). Raising both sides to the positive integer power \(n\) gives \(|a_n|<q^n\), and hence \(|a_n|\leq q^n\). \(\square\)
The proposition provides the central estimate in the convergence case. The Comparison Test for Positive Series compares the tail \(\sum_{n=N}^{\infty}|a_n|\) with the convergent geometric series \(\sum_{n=N}^{\infty}q^n\). The divergence case uses the opposite kind of information: if the root size exceeds one along a subsequence, the corresponding terms cannot approach zero. The Necessary Condition for Series Convergence then rules out convergence.
Connection with the Ratio Test
The two tests often agree when both can be applied. The next result explains why a limit of consecutive ratios controls the nth-root scale for eventually positive terms.
Proof. Fix \(N\geq N_0\). For \(n>N\), repeated multiplication gives $$ a_n=a_N\prod_{k=N}^{n-1}\frac{a_{k+1}}{a_k}. $$
First suppose \(r>0\), and choose \(0<\varepsilon<r\). By the ratio limit, for all sufficiently large \(k\), $$ r-\varepsilon\leq\frac{a_{k+1}}{a_k}\leq r+\varepsilon. $$ Choose \(N\) large enough that these inequalities hold for every \(k\geq N\). Taking the product and then the nth root yields $$ a_N^{1/n}(r-\varepsilon)^{(n-N)/n} \leq a_n^{1/n}\leq a_N^{1/n}(r+\varepsilon)^{(n-N)/n}. $$ As \(n\to\infty\), the left and right sides tend to \(r-\varepsilon\) and \(r+\varepsilon\), respectively, since \(a_N^{1/n}\to1\) and \((n-N)/n\to1\). Therefore the lower and upper limits of \(a_n^{1/n}\) lie between \(r-\varepsilon\) and \(r+\varepsilon\). Allowing \(\varepsilon\) to decrease to zero proves that \(a_n^{1/n}\to r\).
Now suppose \(r=0\). Given any \(\varepsilon>0\), the ratio limit gives an index \(N\) such that \(a_{k+1}/a_k\leq\varepsilon\) for every \(k\geq N\). The product identity implies $$ 0\leq a_n^{1/n}\leq a_N^{1/n}\varepsilon^{(n-N)/n}\qquad(n>N). $$ The right side tends to \(\varepsilon\), so \(\limsup a_n^{1/n}\leq\varepsilon\). Since this holds for every \(\varepsilon>0\), and the roots are nonnegative, \(a_n^{1/n}\to0\). This proves the result in both cases. \(\square\)
This connection explains a useful relationship between the tests, but it does not make them identical. A ratio limit can fail to exist even when the nth-root limit superior is simple, and the Root Test directly measures the terms’ exponential scale without requiring a comparison between consecutive terms.
Worked Applications
Worked Example: An Exponential Denominator with a Polynomial Factor
Consider $$ a_n=\frac{n^2+3}{6^n},\qquad n\geq1. $$ The terms are positive, and $$ |a_n|^{1/n}=\frac{(n^2+3)^{1/n}}{6}. $$ To evaluate the numerator, take logarithms: $$ \log\big((n^2+3)^{1/n}\big)=\frac{\log(n^2+3)}{n}\longrightarrow0. $$ For example, \(n^2+3\leq4n^2\) for \(n\geq2\), so the logarithm divided by \(n\) is bounded above by \((\log4+2\log n)/n\), which tends to zero. Exponentiating shows that \((n^2+3)^{1/n}\to1\). Therefore $$ \lim_{n\to\infty}|a_n|^{1/n}=\frac16<1. $$ The Root Test proves that \(\sum_{n=1}^{\infty}(n^2+3)/6^n\) converges.
Worked Example: Exponential Growth Forces Divergence
Let $$ a_n=\frac{4^n}{n^3+2},\qquad n\geq1. $$ Then $$ |a_n|^{1/n}=\frac{4}{(n^3+2)^{1/n}}. $$ As in the preceding example, $$ \frac{\log(n^3+2)}{n}\longrightarrow0, $$ so \((n^3+2)^{1/n}\to1\). It follows that $$ \lim_{n\to\infty}|a_n|^{1/n}=4>1. $$ The divergence case of the Root Test applies. In particular, the terms do not tend to zero, as is also seen from their exponential growth dominating the polynomial denominator. Thus the series diverges.
Worked Example: The Boundary Case for Two Power Series
For \(a_n=1/n\), the nth root is $$ |a_n|^{1/n}=\exp\left(-\frac{\log n}{n}\right)\longrightarrow1. $$ The harmonic series diverges. For \(b_n=1/n^2\), $$ |b_n|^{1/n}=\exp\left(-\frac{2\log n}{n}\right)\longrightarrow1, $$ but \(\sum_{n=1}^{\infty}1/n^2\) converges by the p-Series Convergence Criterion. Thus the Root Test reaches its boundary value for both series even though their convergence behavior differs. A root limit of one does not decide the series.
Worked Example: An Oscillating Root Sequence
Define positive terms by $$ a_{2k}=\left(\frac13\right)^{2k}, \qquad a_{2k-1}=\left(\frac12\right)^{2k-1} \qquad(k\geq1). $$ Their nth roots alternate: $$ a_{2k}^{1/(2k)}=\frac13, \qquad a_{2k-1}^{1/(2k-1)}=\frac12. $$ The root sequence has no limit, but its limit superior is \(1/2\). Since \(1/2<1\), the Root Test proves convergence. Indeed, the even-indexed terms form a geometric series with ratio \(1/9\), and the odd-indexed terms form a geometric series with ratio \(1/4\): $$ \sum_{k=1}^{\infty}a_{2k}=\sum_{k=1}^{\infty}\left(\frac19\right)^k, \qquad \sum_{k=1}^{\infty}a_{2k-1} =2\sum_{k=1}^{\infty}\left(\frac14\right)^k. $$ Both converge, confirming the Root Test conclusion. This example also shows why the limit-superior formulation is useful when nth roots oscillate.
Interpreting the Test Carefully
A value strictly below one gives a margin: one can choose a fixed \(q<1\) that eventually bounds every nth root, and the terms then lie below \(q^n\). A value strictly above one means that roots exceed some fixed number greater than one infinitely often, so the terms fail the Necessary Condition for Series Convergence. The boundary value one has no such margin in either direction.
The Root Test is particularly effective when terms combine exponential factors and slower-growing factors such as powers of \(n\). Taking nth roots often removes the slower factor and leaves the exponential scale. It can also be more informative than an ordinary root limit: the limit superior still gives an answer when the root sequence oscillates. As with the Ratio Test, however, a boundary value does not settle convergence; another test or a direct argument is then needed.
Check Your Understanding
Use the definition and examples to answer the following questions.
- Why does a limit superior below one allow a geometric bound with ratio strictly below one?
- What does the Root Test conclude if \(\limsup |a_n|^{1/n}=+\infty\)?
- For eventually positive terms with consecutive ratios tending to \(r>0\), what does the nth-root limit equal?
- Why do \(1/n\) and \(1/n^2\) illustrate a limitation of the Root Test?
- In the oscillating example, why is the limit superior more useful than asking whether the nth-root sequence has a limit?