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Root test

The root test, also known as Cauchy's root test, is a criterion in mathematical analysis for assessing the absolute convergence or divergence of an infinite series \sum a_n by evaluating the limit superior of the nth roots of the absolute values of its terms, |a_n|^{1/n}. Introduced by French mathematician Augustin-Louis Cauchy in his 1821 textbook Cours d'analyse, the test provides a systematic way to determine whether the series converges absolutely when this limit is less than 1, diverges when greater than 1, or yields inconclusive results when equal to 1. It is especially effective for series with terms involving exponentials, factorials, or powers, where other tests like the ratio test may be more cumbersome to apply.^[1]^[2]^[3] The precise statement of the root test, valid for series of complex or real numbers, is as follows: Let L = \limsup_{n \to \infty} |a_n|^{1/n}. If L < 1, then \sum |a_n| converges, implying absolute convergence of \sum a_n; if L > 1, then \sum a_n diverges; and if L = 1, the test provides no definitive conclusion, as the series may converge or diverge.^[4] The use of \limsup ensures the test applies even when the ordinary limit \lim_{n \to \infty} |a_n|^{1/n} does not exist, making it a robust tool in real and complex analysis.^[4] Cauchy originally formulated the test in the context of power series and rigorous calculus foundations, emphasizing its role in bounding series behavior through root extraction.^[2] Beyond general series, the root test plays a central role in determining the radius of convergence for power series \sum a_n z^n, where the Cauchy-Hadamard theorem asserts that the radius R satisfies R = 1 / \limsup_{n \to \infty} |a_n|^{1/n}, with R = \infty if the \limsup is 0 and R = 0 if the \limsup is \infty.^[5] This connection highlights the test's foundational importance in analytic function theory and approximation methods, as it directly informs the interval over which the series represents the function.^[5] When inconclusive, the root test is often supplemented by other criteria, such as the comparison test or integral test, to resolve convergence questions.^[3]

Fundamentals of the Root Test

Definition and Criteria

The root test is a criterion for determining the convergence of an infinite series \sum_{n=1}^\infty a_n, where a_n are complex or real numbers. It involves computing the quantity \rho = \limsup_{n \to \infty} |a_n|^{1/n}.^[4] The limit superior is used in the formulation to ensure the test remains applicable even when the ordinary limit \lim_{n \to \infty} |a_n|^{1/n} does not exist, as the lim sup always exists (possibly equal to +\infty).^[4] The convergence criteria of the root test are as follows: if \rho < 1, then the series \sum_{n=1}^\infty a_n converges absolutely; if \rho > 1, then the series diverges; and if \rho = 1, the test provides no definitive conclusion regarding convergence or divergence.^[4] Absolute convergence occurs when the series of absolute values, \sum_{n=1}^\infty |a_n|, converges, which in turn implies that the original series \sum_{n=1}^\infty a_n converges, irrespective of the signs of its terms.^[3] In cases of absolute convergence established by the root test, the series cannot be conditionally convergent, as conditional convergence requires that \sum_{n=1}^\infty a_n converges while \sum_{n=1}^\infty |a_n| diverges.^[3] When \rho = 1, further analysis is needed to distinguish between absolute convergence, conditional convergence, or divergence.^[4] The root test is particularly useful for power series, where \rho determines the reciprocal of the radius of convergence.^[6]

Interpretation of the Limit Superior

The limit superior \rho = \limsup_{n \to \infty} |a_n|^{1/n} in the root test quantifies the asymptotic growth rate of the magnitudes of the series terms |a_n|, providing a measure of how rapidly these terms increase or decrease exponentially as n grows large. This value \rho effectively captures the "envelope" of the term sizes, indicating that for sufficiently large n, |a_n| is bounded above by ( \rho + \epsilon )^n for any \epsilon > 0, akin to the radius of a bounding circle in the complex plane that confines the growth of the terms' magnitudes.^[7] When \rho < 1, the terms |a_n| behave asymptotically like those of a geometric series with common ratio r < 1, which is known to converge, ensuring that the original series converges absolutely by comparison. Conversely, if \rho > 1, the terms grow exponentially, exceeding the growth of a divergent geometric series with ratio r > 1, leading to divergence of the series. This connection to geometric series underscores the root test's reliance on exponential decay or growth as the key determinant of summability.^[7] The use of the limit superior, rather than the plain limit, is crucial for sequences where \lim_{n \to \infty} |a_n|^{1/n} does not exist, such as oscillatory or irregular sequences that fluctuate without settling to a single value. In these cases, \limsup identifies the supremum of the limit points of |a_n|^{1/n}, thereby capturing the worst-case (largest) asymptotic growth rate, which governs the series' convergence behavior even amid oscillations. For instance, in sequences with alternating growth patterns, \limsup ensures the test accounts for subsequences that might otherwise suggest divergence.^[8] The root test was introduced by Augustin-Louis Cauchy in his 1821 textbook Cours d'analyse de l'École Royale Polytechnique, as part of his pioneering work on rigorous criteria for series convergence. The use of the limit superior in the modern formulation ensures applicability even when the ordinary limit does not exist.

Applications of the Root Test

To Infinite Series

The root test is applied to an infinite series \sum a_n by first considering the absolute values of the terms to assess absolute convergence. The practical computation involves evaluating the limit superior \rho = \limsup_{n \to \infty} |a_n|^{1/n}. If \rho < 1, the series converges absolutely (and thus converges); if \rho > 1, the series diverges; and if \rho = 1, the test is inconclusive.^[9]^[3] This test is particularly effective for series involving factorials, exponentials, or super-exponential growth in the terms, where the ratio test may fail to provide a decisive result because the limit of successive term ratios does not exist or is indeterminate. In such cases, the root test leverages the nth root to capture the asymptotic behavior more directly, often yielding a conclusive \rho. For instance, it excels with terms raised to the nth power, simplifying the limit computation compared to ratios.^[9]^[10] Although primarily developed for series with positive terms, the root test extends naturally to arbitrary series by applying it to \sum |a_n|, thereby checking for absolute convergence without addressing conditional convergence. A key limitation arises when \rho = 1, as seen in cases like the harmonic series, where the test provides no information on convergence or divergence, necessitating alternative tests.^[3]^[11]

To Power Series

The root test extends naturally to power series of the form \sum_{n=0}^{\infty} c_n (z - a)^n, where z is a complex variable and a is the center, to determine the disk of convergence in the complex plane. By applying the test to the absolute values of the terms, the radius of convergence R is given by R = \frac{1}{\rho}, where \rho = \limsup_{n \to \infty} |c_n|^{1/n}.^[12] This formula, known as the Cauchy-Hadamard theorem, provides a precise measure of the series' convergence behavior.^[13] Within this radius, the power series converges absolutely for all z satisfying |z - a| < R, forming an open disk centered at a. Conversely, the series diverges for all |z - a| > R. At the boundary points where |z - a| = R, convergence must be assessed separately using other tests, as the root test yields no definitive conclusion there.^[14] Special cases arise depending on the value of \rho: if \rho = 0, then R = \infty, and the series converges everywhere in the complex plane (the entire plane); if \rho = \infty, then R = 0, and the series converges only at the center z = a. The root test formulation is particularly advantageous for determining the radius of convergence in cases involving analytic functions, as it aligns directly with Hadamard's formula and applies even when the ratio test limit does not exist.^[15]^[13]

Proof of the Root Test

Convergence Implication

The convergence implication of the root test states that if \rho = \limsup_{n \to \infty} |a_n|^{1/n} < 1 for a series \sum a_n, then the series \sum |a_n| converges, implying absolute convergence of \sum a_n.^[16]^[3] To prove this, assume \rho < 1. Select a real number r such that \rho < r < 1. By the definition of the limit superior, there exists a positive integer N such that for all n > N, |a_n|^{1/n} < r. Raising both sides to the power n yields |a_n| < r^n for all n > N.^[16]^[3] The tail of the series \sum_{n=N+1}^\infty |a_n| can now be bounded above by the geometric series \sum_{n=N+1}^\infty r^n. Specifically, |a_n| < r^n for n > N, and the geometric series \sum_{n=0}^\infty r^n = \frac{1}{1-r} converges since $0 < r < 1. By the comparison test, \sum_{n=N+1}^\infty |a_n| converges because each term is less than or equal to the corresponding term of a convergent series. The partial sum \sum_{n=1}^N |a_n| is finite, so the full series \sum_{n=1}^\infty |a_n| converges.^[16]^[3] Thus, \sum a_n converges absolutely.^[16]^[3]

Divergence Implication

The divergence implication of the root test states that if \rho = \limsup_{n \to \infty} |a_n|^{1/n} > 1, then the series \sum a_n diverges.^[17] To establish this, recall the definition of the limit superior: \rho = \inf_{n \geq 1} \sup_{k \geq n} |a_k|^{1/k}. Since \rho > 1, there exists \varepsilon > 0 such that \rho > 1 + \varepsilon, ensuring that \sup_{k \geq n} |a_k|^{1/k} > 1 + \varepsilon for every n. Consequently, for each n, there is at least one k_n \geq n satisfying |a_{k_n}|^{1/k_n} > 1 + \varepsilon. This constructs a subsequence \{k_n\} with k_n \to \infty as n \to \infty.^[18] Along this subsequence, |a_{k_n}| > (1 + \varepsilon)^{k_n}. Since $1 + \varepsilon > 1, the right-hand side (1 + \varepsilon)^{k_n} \to \infty as n \to \infty, implying |a_{k_n}| \to \infty. Thus, the terms a_n do not converge to 0, as the subsequence diverges to infinity.^[17] By the term test for divergence, if a_n \not\to 0, the series \sum a_n cannot converge and must diverge. This completes the proof, highlighting how \rho > 1 prevents the necessary condition for convergence.^[19]

Illustrative Examples

Convergent Cases

One illustrative example of the root test applied to a convergent series involves the power series \sum_{n=0}^{\infty} \frac{n!}{n^n} x^n evaluated at x=1, yielding the numerical series \sum_{n=0}^{\infty} \frac{n!}{n^n}. Here, the general term is a_n = \frac{n!}{n^n}, so |a_n|^{1/n} = \frac{(n!)^{1/n}}{n}. Using Stirling's approximation n! \approx \sqrt{2\pi n} (n/e)^n, it follows that (n!)^{1/n} \approx (2\pi n)^{1/(2n)} \cdot (n/e), and as n \to \infty, (2\pi n)^{1/(2n)} \to 1, yielding \lim_{n \to \infty} |a_n|^{1/n} = 1/e \approx 0.3679 < 1. Thus, the root test implies absolute convergence of the series. Another classic convergent case is the exponential power series \sum_{n=0}^{\infty} \frac{x^n}{n!}, where the general term is a_n = \frac{x^n}{n!} and \rho = \limsup_{n \to \infty} |a_n|^{1/n} = \lim_{n \to \infty} \frac{|x|}{(n!)^{1/n}}. Applying Stirling's approximation again, (n!)^{1/n} \sim n/e, so \frac{|x|}{(n!)^{1/n}} \sim |x| \cdot \frac{e}{n} \to 0 < 1 for any fixed x \in \mathbb{R}. Therefore, \rho = 0 < 1, and the root test confirms absolute convergence for all x, establishing an infinite radius of convergence. These examples highlight the root test's utility for series involving factorial terms, where the test directly extracts the limiting behavior through the nth root, often yielding the same convergence limit as the ratio test but providing a straightforward approach for terms with exponents or factorials.

Divergent and Inconclusive Cases

In cases where the limit superior \rho = \limsup_{n \to \infty} |a_n|^{1/n} > 1, the root test establishes divergence of the series \sum a_n, as the terms do not approach zero sufficiently rapidly.^[3] For instance, consider the series \sum_{n=1}^\infty \frac{n^n}{n!}. Here, |a_n|^{1/n} = \frac{n}{(n!)^{1/n}}, and using Stirling's approximation, (n!)^{1/n} \sim \frac{n}{e}, so the limit is e \approx 2.718 > 1. Thus, the series diverges by the root test. A stronger form of divergence occurs when \rho = \infty > 1. An example is the series \sum_{n=1}^\infty n!, where a_n = n! and |a_n|^{1/n} = (n!)^{1/n} \sim \frac{n}{e} \to \infty. This implies the terms grow too rapidly for convergence, confirming divergence. The root test is inconclusive when \rho = 1, providing no definitive conclusion about convergence or divergence. A classic example is the harmonic series \sum_{n=1}^\infty \frac{1}{n}, which diverges but yields \rho = 1. Specifically, |a_n|^{1/n} = \left(\frac{1}{n}\right)^{1/n} = \frac{1}{n^{1/n}}, and since n^{1/n} \to 1, the limit is $1. In such cases, alternative tests like the integral test are required, which shows divergence via \int_1^\infty \frac{1}{x} \, dx = \infty.^[20] For power series, the root test determines the radius of convergence R = \frac{1}{\rho}. Consider \sum_{n=0}^\infty n! z^n; here, \rho = \limsup_{n \to \infty} (n!)^{1/n} = \infty, so R = 0, and the series diverges for all z \neq 0. This aligns with the application of the root test to power series, where such a radius indicates convergence only at the center.

Comparisons and Extensions

With the Ratio Test

The ratio test assesses the convergence of an infinite series \sum a_n by evaluating L = \limsup_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|; the series converges absolutely if L < 1, diverges if L > 1, and the test is inconclusive if L = 1.(https://openstax.org/books/calculus-volume-2/pages/9-6-the-ratio-and-root-tests) This test is particularly effective for series involving factorials or exponentials, where the recursive structure simplifies the computation of successive term ratios.(https://openstax.org/books/calculus-volume-2/pages/9-6-the-ratio-and-root-tests) The root test and ratio test exhibit a close relationship, as for many series where both limits exist, \lim_{n \to \infty} |a_n|^{1/n} equals \lim_{n \to \infty} |a_{n+1}/a_n|.^[21] More generally, \liminf_{n \to \infty} |a_{n+1}/a_n| \leq \liminf_{n \to \infty} |a_n|^{1/n} \leq \limsup_{n \to \infty} |a_n|^{1/n} \leq \limsup_{n \to \infty} |a_{n+1}/a_n|, showing that the root test encompasses the ratio test's conclusions.^[21] In the context of power series, the Cauchy-Hadamard theorem establishes the radius of convergence R as R = 1 / \limsup_{n \to \infty} |a_n|^{1/n}, directly employing the root test, while the ratio test yields R = 1 / \lim_{n \to \infty} |a_{n+1}/a_n| when that limit exists, thereby linking the two for determining the interval of convergence.^[15] A key difference is that the root test is stronger for irregular series where the ratios fluctuate, as it relies on the geometric mean of the terms rather than consecutive pairwise comparisons. For instance, consider the series with terms defined such that the ratios |a_{n+1}/a_n| alternate between 2 and $1/8; the limit of the ratios does not exist, rendering the ratio test inconclusive, but the root test gives \lim_{n \to \infty} |a_n|^{1/n} = 1/2 < 1, confirming absolute convergence.^[21] Conversely, the ratio test is often preferable for series with terms that naturally lend themselves to recursive evaluation, such as those featuring factorials. Both tests produce the same radius of convergence for power series when the ratio limit exists, but the root test's reliance on \limsup provides greater robustness in cases of non-existent ratio limits.^[15]

Advanced Variants and Hierarchy

Advanced variants of the root test extend its applicability to series where the standard criterion yields inconclusive results, particularly when the limit equals 1. One such extension involves higher-order root tests, which refine the analysis by considering slower-growing exponents in the root. For instance, the test examines \limsup_{n \to \infty} |a_n|^{1/(n \log n)}; if this value is less than 1, the series converges absolutely, while greater than 1 implies divergence. This variant addresses cases of superlinear growth where the basic root test fails to distinguish convergence.^[22] Raabe's test serves as a hybrid between root and ratio approaches, enhancing precision for borderline series. It states that for positive terms, if \lim_{n \to \infty} n \left( \frac{a_n}{a_{n+1}} - 1 \right) > 1, the series converges, and if less than 1, it diverges; the case of equality is inconclusive. Developed by Joseph Ludwig Raabe in the 19th century, this test combines the multiplicative nature of the root test with the sequential comparison of the ratio test, making it effective for p-series refinements.^[23] A notable generalization is de la Vallée Poussin's criterion for mean roots, which further broadens the root test's scope by incorporating logarithmic adjustments to the root index, such as considering terms like I_n - \ln(\ln n)/\ln n where I_n involves logarithmic ratios of consecutive terms. Introduced by Charles-Jean de la Vallée Poussin in the early 20th century, this variant applies to series like \sum \frac{1}{n (\ln n)^k} for k > 1, where it confirms convergence beyond the standard root test's reach.^[22] Within the hierarchy of convergence tests, the root test occupies a mid-level position, surpassing the ratio test in handling superlinear growth patterns, such as factorial or exponential terms where ratios oscillate but roots stabilize below 1. However, it falls short of the integral test for cases where the root limit \rho = 1, requiring integral comparisons for logarithmic or polynomial borderline series. As part of Cauchy's broader framework of root-based criteria, it forms the foundation for subsequent refinements like Raabe's and de la Vallée Poussin's, creating an ascending chain of increasingly powerful tests.^[22] In practical application, the root test follows simpler diagnostics like the term test for divergence and precedes advanced tools for conditional convergence, such as Dirichlet's test, ensuring a logical progression in analyzing infinite series.^[22]

References

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Root Test -- from Wolfram MathWorld
1. If rho<1 , the series converges. 2. If rho>1 or rho=infty , the series diverges. 3. If rho=1 , the series may converge or diverge. This test is also called ...
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Cours d'analyse de l'Ecole royale polytechnique - Internet Archive
May 1, 2016 · Cours d'analyse de l'Ecole royale polytechnique; par m. AugustinLouis Cauchy ... 1.re partie. Analyse algébrique ; Publication date: 1821 ; Usage ...
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Calculus II - Root Test - Pauls Online Math Notes
Nov 16, 2022 · In this section we will discuss using the Root Test to determine if an infinite series converges absolutely or diverges. The Root Test can ...
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Root Test - Real Analysis - MathCS.org
It is important to remember that when the root test gives 1 as the answer for the lim sup, then no conclusion at all is possible. The use of the lim sup rather ...
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Cauchy-Hadamard Theorem -- from Wolfram MathWorld
### Summary of Cauchy-Hadamard Theorem
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Power Series
So the radius of convergence is 1L and the interval of convergence is from −1L to 1L. The same logic holds for the Root Test.
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[PDF] ∑ ∑ ∑ ∑ ∑kan = lim ∑ ∑ ∑ aj = k ∑an )(∑bn - UCI Mathematics
Definition 3.3 (Geometric series). A sequence (an) is geometric if the ... Theorem 3.15 (Root test). Suppose lim sup |an|. 1/n. = L. 1. If L < 1, then ...
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[PDF] Supplement on lim sup and lim inf
anbn ≤ lim sup n→ an lim sup n→ bn . Remark: (1) The equality ... root test is more powerful than the ratio test. We give an example to say ...
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[PDF] RES.18-001 Calculus (f17), Chapter 10: Infinite Series
EXAMPLE 8 The series P1=nn is ideal for the root test. The nth root is 1=n: Its limit is L D 0: Convergence is even faster than for e D P1=nŠ The root test ...Missing: criteria | Show results with:criteria
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Although this Root Test is more difficult to apply, it is better than the Ratio Test in the following sense. There are series for which the Ratio Test give no ...
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[PDF] Limits and Infinite Series Branko´Curgus
Oct 2, 2020 · (b) If R > 1, then the series diverges. Notice that if the root or the ratio test apply to a series, then series either converges absolutely ...
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[PDF] Chapter 10: Power Series - UC Davis Math
The root test gives an expression for the radius of convergence of a general power series. Theorem 10.6 (Hadamard). The radius of convergence R of the power ...
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be used to determine the radius of convergence. The root test yields 1/R = limn→∞ |an|1/n. If the latter fails to exist, one can modify ...
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None
- **Cauchy-Hadamard Formula**: The radius of convergence $ R $ of the power series $ \sum_{n=0}^{\infty} c_n (z - z_0)^n $ is $ R = \frac{1}{\lim_{n \to \infty} \sqrt[n]{|c_n|}} $.
[15]
[PDF] Power Series - Trinity University
Hadamard's Formula. Examples. Example 2. Determine the radius of convergence of. ∞. X k=1 zk k2k . Solution. We appeal to the root test for absolute convergence ...
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Mar 30, 2016 · In this section, we prove the last two series convergence tests: the ratio test and the root test. These tests are particularly nice because ...
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[PDF] Math 4111 October 9, 2020 Lecture
Oct 9, 2020 · Now we have the first nontrivial convergence test for series. Theorem (The Root Test). Consider the series P. ∞ j=1 aj and let r = lim sup j ...
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Question about proof of root test for $\alpha=\lim \sup a_{n}>1
Jan 23, 2020 · Question about proof of root test for α=limsupan>1. Given ∑∞n=1an let limsup(an)1/n=α∈R. a) If α<1, then ∑∞n=1an converges. b) If α>1 then ∑∞n= ...Root test and $\lim\sup$ - Mathematics Stack Exchangeroot test: why $\lim\sup$? - Mathematics Stack ExchangeMore results from math.stackexchange.com
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None
### Summary of Proof for Root Test Case (b): α > 1, Series Diverges
[20]
More Challenging Problems: The ratio and root tests
Jul 6, 2017 · 1. The presence of n in exponents suggests the Root Test. The trick is to recognize the denominator is 2n(2)= (2n)n, so an can be rewritten as ( ...<|control11|><|separator|>
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[PDF] Math 131Infinite Series, Part V: The Ratio and Root Tests
Why is the test inconclusive when the root is 1? The next example shows why. EXAMPLE 14.35. Consider the harmonic series. ∞. ∑.
[22]
[PDF] The Relation Between the Root and Ratio Tests | USC Dornsife
then the series Can converges. This test is stronger than the ratio test but weaker than the root test. However, in some cases it may be easier ...Missing: advantages | Show results with:advantages
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[PDF] A Hierarchy of the Convergence Tests Related to Cauchy's Test
In this study we apply a general theorem on convergence of numerical positive series in order to construct a hierarchy of the specific tests intimately.
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May 8, 2019 · Abstract. The primary objective of this paper is to discuss advanced tests of convergence for infinite series.<|control11|><|separator|>