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Ratio test

The ratio test is a criterion in calculus for determining the convergence or divergence of an infinite series \sum a_n, where a_n are the terms of the series. It involves computing the limit L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|; if L < 1, the series converges absolutely (and hence converges); if L > 1, the series diverges; and if L = 1, the test is inconclusive, requiring other methods to assess convergence.^[1]^[2]^[3] This test is particularly effective for series involving exponentials, factorials, or powers, such as the exponential series \sum \frac{x^n}{n!} or power series, where the ratio of successive terms simplifies to a clear limit.^[4]^[5] Developed in the 18th century by Jean le Rond d'Alembert, alongside other tests for infinite series such as the root test (later formalized by Cauchy), it provides a practical tool for absolute convergence without relying on integral or comparison tests in many cases. When the limit L = 1 occurs, as in the harmonic series \sum \frac{1}{n}, alternative criteria like the integral test or p-series test must be applied to resolve the behavior.^[6]^[7]

Introduction

Definition and Purpose

The ratio test serves as a fundamental tool in calculus for assessing the absolute convergence of an infinite series \sum a_n by evaluating the limit L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|.^[8]^[9] This approach focuses on the behavior of the ratio of consecutive terms to infer whether the series converges in the sense that the sum of the absolute values remains finite, thereby guaranteeing convergence regardless of the signs of the terms.^[9] To apply the ratio test effectively, one must possess a foundational understanding of infinite series, the notion of absolute convergence, and the computation of limits for sequences.^[9] These prerequisites ensure that the test can be used to analyze the tail of the series, where the limit determines the overall behavior. The primary purpose of the ratio test is to provide a straightforward method for convergence analysis, particularly for series featuring factorials, exponentials, or powers, as the ratios of successive terms often simplify dramatically, yielding computable limits.^[8]^[9] It is preferred over the comparison test in such cases because the latter demands identifying a benchmark series with known convergence properties, which can be challenging for complex expressions, whereas the ratio test operates directly on the given series' terms.^[9] As one of several limit-based convergence tests, the ratio test shares conceptual similarities with the root test, which instead examines the limit of the nth root of the absolute term.^[9]

Historical Background

The ratio test emerged in the context of the 18th-century efforts to establish rigorous criteria for the convergence of infinite series, building on the foundational work in calculus by Isaac Newton and Gottfried Wilhelm Leibniz earlier that century. As mathematicians grappled with the summation of series beyond simple geometric progressions—where convergence is determined by the common ratio being less than 1 in absolute value—early tests sought to generalize this ratio-based intuition to broader classes of series.^[10] The test was first formally published by Jean le Rond d'Alembert in 1768, in the fifth volume of his Opuscules mathématiques, earning it the name d'Alembert's ratio test. D'Alembert's contribution provided an initial criterion for assessing series convergence by examining the limit of the ratio of successive terms, marking a significant step in the systematic study of infinite series during the Enlightenment era.)^[11] In the early 19th century, Augustin-Louis Cauchy reformulated and rigorously proved the test in his seminal textbook Cours d'analyse de l'École Royale Polytechnique (1821), integrating it into a comprehensive framework for real analysis. This version, often called Cauchy's ratio test, emphasized strict limits and absolute convergence, reflecting Cauchy's push for epsilon-delta precision in calculus and influencing the development of subsequent convergence criteria.^[12]^[13] Augustus De Morgan extended the ratio test in the mid-19th century by developing a hierarchy of refined tests to handle inconclusive cases, as detailed in his works on series convergence; these extensions, later linked with Joseph Bertrand's contributions, enhanced the test's applicability in mathematical analysis.^[14]

Formal Statement

The Theorem

The ratio test provides a criterion for determining the convergence or divergence of an infinite series based on the limit of the ratio of consecutive terms. Formally stated, consider the series \sum_{n=1}^\infty a_n, where the terms a_n are complex numbers with a_n \neq 0 for all sufficiently large n. Define

L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|,

assuming this limit exists.^[15] If L < 1, then the series \sum |a_n| converges, implying that \sum a_n converges absolutely (and thus converges). If L > 1, then the terms |a_n| do not approach zero, so the series \sum a_n diverges. If L = 1, the test yields no information about convergence or divergence, as the series may either converge or diverge.^[15]/08%3A_Sequences_and_Series/8.06%3A_The_Ratio_Test) This theorem assumes the existence of the limit L; if the limit does not exist, the test cannot be applied directly. The use of absolute values in the ratio ensures the criterion applies to series with possibly negative or complex terms, with the focus on absolute convergence when L < 1. For series of positive real terms, the absolute value notation is often omitted, but the conclusions remain analogous for ordinary convergence.^[16]^[15] The ratio test was originally formulated by Jean le Rond d'Alembert in 1768, appearing in his work Opuscules mathématiques.^[16]

Interpretation of the Limit L

The limit L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| in the ratio test reveals the asymptotic growth or decay rate of the series terms relative to a geometric progression. When L < 1, the terms |a_n| diminish faster than those in a geometric series with common ratio r where |r| < 1, ensuring the partial sums are bounded and leading to absolute convergence of the series.^[9] In contrast, if L > 1, the magnitudes of the terms increase exponentially, akin to a geometric series with |r| > 1, which results in unbounded partial sums and divergence of the series.^[1] The case L = 1 provides no definitive conclusion, as the test cannot distinguish between convergence and divergence; additional tests are required to assess the series behavior.^[2] Should the limit L fail to exist, the ratio test cannot be applied, necessitating the use of other convergence criteria to evaluate the series.^[2]

Proof

Proof for Absolute Convergence (L < 1)

Assume the limit L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| exists and satisfies L < 1, where \{a_n\} is a sequence of nonzero terms.^[17]^[18] Since L < 1, there exists a real number r such that L < r < 1. By the definition of the limit, there is some integer N such that for all n \geq N,

\left| \frac{a_{n+1}}{a_n} \right| < r.

This inequality implies |a_{n+1}| < r |a_n| for n \geq N.^[17]^[18] To bound the terms |a_n| for n > N, proceed by induction. For the base case k = 1, |a_{N+1}| < r |a_N|. Assume the statement holds for some k = m \geq 1, so |a_{N+m}| < r^m |a_N|. Then for k = m+1,

|a_{N+m+1}| < r |a_{N+m}| < r (r^m |a_N|) = r^{m+1} |a_N|,

which completes the induction. Thus, for all k \geq 1,

|a_{N+k}| < |a_N| r^k.

^[18] Consider the tail of the absolute series starting from n = N+1:

\sum_{k=1}^\infty |a_{N+k}| < \sum_{k=1}^\infty |a_N| r^k = |a_N| \sum_{k=1}^\infty r^k.

The geometric series \sum_{k=1}^\infty r^k converges to \frac{r}{1-r} since $0 \leq r < 1, so

\sum_{k=1}^\infty |a_{N+k}| < |a_N| \frac{r}{1-r},

a finite bound. By the comparison test, the tail \sum_{n=N+1}^\infty |a_n| converges.^[17]^[18] The full series \sum_{n=1}^\infty |a_n| is then the sum of the finite partial sum \sum_{n=1}^N |a_n| and the convergent tail, hence converges. Therefore, \sum_{n=1}^\infty a_n converges absolutely.^[17]^[18]

Proof for Divergence (L > 1)

To prove that the series \sum a_n diverges when L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| > 1, it suffices to show that the terms a_n do not approach zero as n \to \infty, which violates the necessary condition for convergence of any series.^[19]^[1] Since L > 1, there exists some integer N such that for all n \geq N, \left| \frac{a_{n+1}}{a_n} \right| > 1. More precisely, choose \epsilon = \frac{L-1}{2} > 0, so that $1 + \epsilon = \frac{L+1}{2} satisfies $1 < 1 + \epsilon < L. Then, for all n \geq N, \left| \frac{a_{n+1}}{a_n} \right| > 1 + \epsilon, which implies \left| a_{n+1} \right| > (1 + \epsilon) \left| a_n \right|.^[1]^[2] Iterating this inequality for k \geq 0, it follows that \left| a_{N+k} \right| > (1 + \epsilon)^k \left| a_N \right|. Since $1 + \epsilon > 1, the right-hand side grows exponentially to infinity as k \to \infty, so \left| a_n \right| is bounded below by a positive constant c = \left| a_N \right| for infinitely many n (in fact, for all sufficiently large n), and thus \lim_{n \to \infty} a_n \neq 0.^[19]^[1] Therefore, the nth-term test for divergence applies: if \lim_{n \to \infty} a_n \neq 0, the series \sum a_n diverges.^[19]^[2]

Examples

Convergent Series (L < 1)

One prominent example of a convergent series identified by the ratio test is the geometric series \sum_{n=0}^{\infty} r^n, where |r| < 1. The general term is a_n = r^n. The ratio of consecutive terms is \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{r^{n+1}}{r^n} \right| = |r|. Taking the limit as n \to \infty yields L = |r|. Since L < 1, the ratio test concludes that the series converges absolutely.^[20] Another illustrative case is the exponential series \sum_{n=0}^{\infty} \frac{x^n}{n!}, which converges for all real x. Here, the general term is a_n = \frac{x^n}{n!}. The ratio is \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{x^{n+1}/(n+1)!}{x^n/n!} \right| = \frac{|x|}{n+1}. The limit as n \to \infty is L = \lim_{n \to \infty} \frac{|x|}{n+1} = 0 < 1. Thus, the ratio test confirms absolute convergence for any x.^[21] The series \sum_{n=1}^{\infty} \frac{n!}{n^n} also demonstrates convergence via the ratio test. The general term is a_n = \frac{n!}{n^n}. The ratio of consecutive terms is \left| \frac{a_{n+1}}{a_n} \right| = \frac{(n+1)!/(n+1)^{n+1}}{n!/n^n} = \frac{(n+1) n! \cdot n^n}{n! \cdot (n+1)^{n+1}} = \frac{n^n}{(n+1)^n} = \left( \frac{n}{n+1} \right)^n. The limit is L = \lim_{n \to \infty} \left( \frac{n}{n+1} \right)^n = \lim_{n \to \infty} \left( 1 - \frac{1}{n+1} \right)^n = e^{-1} \approx 0.3679 < 1. Therefore, the series converges absolutely.^[22]

Divergent Series (L > 1)

When the limit L > 1 in the ratio test, the series diverges because the terms fail to approach zero sufficiently rapidly, as established by the theorem's divergence criterion.^[1] A classic example is the geometric series \sum_{n=0}^{\infty} 2^n. The general term is a_n = 2^n, so the ratio is

\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{2^{n+1}}{2^n} = 2 > 1.

Thus, L = 2 > 1, and the series diverges.^[20] Another illustrative case is the factorial series \sum_{n=1}^{\infty} n!. Here, a_n = n!, and the ratio simplifies to

\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{(n+1)!}{n!} = \lim_{n \to \infty} (n+1) = \infty > 1.

Since L = \infty > 1, the series diverges.^[23] For a more rapidly growing example, consider \sum_{n=1}^{\infty} n^n. The terms are a_n = n^n, yielding the ratio

\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{(n+1)^{n+1}}{n^n} = \lim_{n \to \infty} (n+1) \left(1 + \frac{1}{n}\right)^n = \infty \cdot e = \infty > 1.

With L = \infty > 1, this series also diverges.^[1]

Inconclusive Cases (L = 1)

When the limit L = 1, the ratio test provides no definitive conclusion about the convergence or divergence of the series, necessitating the use of alternative tests.^[24] Consider the harmonic series \sum_{n=1}^\infty \frac{1}{n}. Here, a_n = \frac{1}{n}, so

L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{n}{n+1} = 1.

The series diverges, as established by the integral test applied to f(x) = \frac{1}{x}, where \int_1^\infty \frac{1}{x} \, dx = \infty.^[25] Next, examine the alternating harmonic series \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}. The terms are a_n = \frac{(-1)^{n+1}}{n}, and the ratio test considers the absolute values, yielding

L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{n}{n+1} = 1.

This series converges conditionally by the alternating series test but does not converge absolutely, since the absolute series is the divergent harmonic series.^[26] Finally, consider the p-series \sum_{n=1}^\infty \frac{1}{n^2}. With a_n = \frac{1}{n^2},

L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{n^2}{(n+1)^2} = 1.

The series converges, as shown by the integral test on f(x) = \frac{1}{x^2}, where \int_1^\infty \frac{1}{x^2} \, dx = 1 < \infty.^[25]

Applications

Power Series and Radius of Convergence

The ratio test provides a powerful method for determining the radius of convergence of a power series \sum_{n=0}^\infty c_n (x - a)^n. Applying the test to the absolute value of consecutive terms yields \lim_{n \to \infty} \left| \frac{c_{n+1} (x - a)^{n+1}}{c_n (x - a)^n} \right| = \left| x - a \right| \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right|. If the limit L = \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right| exists and is finite, the series converges absolutely when L |x - a| < 1, or equivalently, when |x - a| < 1/L. Thus, the radius of convergence is R = 1/L.^[27] Equivalently, if the limit \lim_{n \to \infty} \left| \frac{c_n}{c_{n+1}} \right| exists, then R equals this value, providing an alternative formula for computation.^[28] For the geometric series \sum_{n=0}^\infty x^n, the coefficients are c_n = 1, so L = \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right| = 1, yielding R = 1. The series converges for |x| < 1 and diverges for |x| > 1.^[29] In contrast, the Taylor series for e^x, \sum_{n=0}^\infty \frac{x^n}{n!}, has c_n = 1/n!, so L = \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right| = \lim_{n \to \infty} \frac{1}{n+1} = 0, implying R = \infty and convergence for all x.^[28] The interval of convergence consists of all x such that |x - a| < R, with divergence guaranteed for |x - a| > R. When R is finite, the behavior at the endpoints x = a \pm R (where |x - a| = R) must be tested separately using other convergence criteria, as the ratio test is inconclusive there.^[27]

Exponential and Factorial Series

The exponential series \sum_{n=0}^{\infty} \frac{x^n}{n!} provides a fundamental application of the ratio test to factorial-denominated terms. To assess convergence, compute the limit L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|, where a_n = \frac{x^n}{n!}. This yields L = \lim_{n \to \infty} \frac{|x|^{n+1}/(n+1)!}{|x|^n/n!} = \lim_{n \to \infty} \frac{|x|}{n+1} = 0 < 1 for any fixed x \in \mathbb{R}. Since L < 1, the series converges absolutely for all real x, establishing the exponential function e^x as an entire function with infinite radius of convergence.^[30] The rapid decay of terms, driven by the superlinear growth of n!, ensures that partial sums approximate e^x efficiently even for moderate n, with the error bounded by the next term in the series.^[1] Similar behavior arises in the Taylor series for sine and cosine, which feature factorials in the denominators alongside alternating signs and odd or even powers. For \sin x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}, the ratio test gives L = \lim_{n \to \infty} \frac{|x|^{2(n+1)+1}/(2(n+1)+1)!}{|x|^{2n+1}/(2n+1)!} = \lim_{n \to \infty} \frac{|x|^2}{(2n+2)(2n+3)} = 0 < 1, confirming absolute convergence for all x. Analogously, for \cos x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!}, L = \lim_{n \to \infty} \frac{|x|^{2(n+1)}/(2(n+1))!}{|x|^{2n}/(2n)!} = \lim_{n \to \infty} \frac{|x|^2}{(2n+1)(2n+2)} = 0 < 1. These results underscore the entire nature of \sin x and \cos x, with factorial growth enabling rapid convergence and high-fidelity approximations in computational contexts.^[31] Bessel functions of the first kind exemplify how the ratio test simplifies for special functions with recursive factorial-like coefficients. The series J_\nu(x) = \left( \frac{x}{2} \right)^\nu \sum_{k=0}^{\infty} \frac{(-1)^k}{k! \, \Gamma(\nu + k + 1)} \left( \frac{x}{2} \right)^{2k} involves gamma functions that generalize factorials. Applying the ratio test to the terms b_k = \frac{(-1)^k}{k! \, \Gamma(\nu + k + 1)} \left( \frac{x}{2} \right)^{2k}, the limit \rho = \lim_{k \to \infty} \left| \frac{b_{k+1}}{b_k} \right| = \lim_{k \to \infty} \frac{|x|^2 / 4}{(k+1)(\nu + k + 1)} = 0 < 1 for any fixed x > 0 and order \nu. This establishes convergence for all x > 0, highlighting the test's utility for series with recursive coefficients in differential equations.^[32]^[33] For factorial series where the direct ratio test may yield L = 1 at the boundary of convergence, Stirling's approximation n! \sim \sqrt{2 \pi n} \, (n/e)^n refines the analysis of term behavior. Consider series like \sum_{n=1}^{\infty} \frac{n! \, x^n}{n^{n}}; the ratio \left| \frac{a_{n+1}}{a_n} \right| = |x| \left( \frac{n}{n+1} \right)^n \to |x| / e. Thus, the radius of convergence is e, with absolute convergence for |x| < e. At the boundary |x| = e, L = 1 and the test is inconclusive, but Stirling shows the terms a_n \sim \sqrt{2 \pi n}, which tend to infinity and do not approach zero, implying divergence. The approximation's accuracy for large n quantifies the rapid term reduction in factorial-dominated series within the radius of convergence.^[34]

Extensions for L = 1

De Morgan's Hierarchy

In the 19th century, British mathematician Augustus De Morgan contributed to the development of higher-order ratio tests to resolve inconclusive cases of the basic ratio test, where the limit L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 1. This work built upon earlier contributions by d'Alembert and Cauchy, aiding in the systematic analysis of series convergence when the standard test is indeterminate. De Morgan's efforts helped establish a progression of refinements for borderline series.^[35] The hierarchy starts with the basic d'Alembert ratio test as the first level (order 1), examining L directly. Higher orders incorporate asymptotic expansions of the ratio. For example, the second order may involve limits like \lim_{n \to \infty} n \left( 1 - \left| \frac{a_n}{a_{n+1}} \right| \right), with generalizations for subsequent levels. Each level decides convergence or divergence where prior levels are inconclusive, such as harmonic-like series at order 2 or those with logarithmic terms at higher orders. The basic ratio test forms level 1 in this framework.^[36]^[37] In general, the k-th order test evaluates expressions like \lim_{n \to \infty} n^k \left( \frac{a_n}{a_{n+1}} - 1 \right), where the series typically converges if this limit exceeds 1 and diverges if below 1, with inconclusiveness leading to order k+1. Higher orders may include iterated logarithms, such as \frac{1}{n \ln n} or \frac{1}{n \ln n \ln \ln n}, for slowly varying functions. De Morgan's contributions to analysis in the 19th century helped lay the foundation for this progression, influencing later work by Bertrand, Kummer, and others.^[14]^[36] Several advanced refinements to the ratio test address cases where the standard test yields an inconclusive result (L=1), providing more precise criteria for convergence or divergence of series with positive terms. These tests build upon foundational ideas in De Morgan's hierarchy by incorporating higher-order asymptotic behaviors or generalized comparisons.^[38] Raabe's test, proposed by Joseph Ludwig Raabe in 1832, refines the ratio by multiplying the difference from unity by n. Specifically, for a series \sum a_n with a_n > 0, compute L = \lim_{n \to \infty} n \left( \frac{a_n}{a_{n+1}} - 1 \right). The series converges if L > 1 and diverges if L < 1; if L = 1, the test is inconclusive.^[39] This test succeeds where the basic ratio test fails, such as for the series \sum 1/\sqrt{n}, where Raabe's L = 1/2 < 1 correctly indicates divergence. Bertrand's test extends Raabe's approach by accounting for a logarithmic factor, applicable when Raabe's limit equals 1. The condition is L = \lim_{n \to \infty} \ln n \left[ n \left( \frac{a_n}{a_{n+1}} - 1 \right) - 1 \right]; the series converges if L > 1 and diverges if L < 1.^[40] For example, consider \sum_{n=3}^\infty \frac{1}{n (\ln n)^2}, where both the basic ratio and Raabe's tests are inconclusive (L=1), but Bertrand's L = 2 > 1 confirms convergence.^[41] Gauss's test examines the asymptotic expansion of the ratio, assuming \frac{a_{n+1}}{a_n} = 1 - \frac{b}{n} + o\left(\frac{1}{n}\right) for some constant b. The series converges if b > 1 and diverges if 0 < b ≤ 1 (assuming the o(1/n) term allows the expansion).^[42] This applies when lower-order tests fail, as in certain hypergeometric series where the basic ratio gives L=1 and Raabe is inconclusive, but Gauss's b>1 establishes convergence. Kummer's test, introduced by Ernst Kummer in 1835, generalizes ratio comparisons by involving a known convergent or divergent series \sum b_n with b_n > 0. Define s_n = \prod_{k=1}^n \frac{b_k}{a_k}; if \liminf_{n \to \infty} s_n \frac{a_n}{b_n} > 0 and \sum b_n converges, then \sum a_n converges; if the lim sup < 0 or \sum b_n diverges under similar conditions, \sum a_n diverges.^[43] An example is testing \sum_{n=2}^\infty \frac{1}{n^2 \ln n} (for n≥2) against the known convergent p-series \sum \frac{1}{n^2} (b_n = 1/n^2), where the basic ratio fails (L=1), but Kummer's lim inf >0 implies convergence.^[44] Tong's modification of Kummer's test, developed by Jingcheng Tong in 1994, provides a characterization that covers all positive-term series by ensuring the test can determine convergence or divergence definitively through appropriate choice of {p_n} in the product form, extending Kummer to exhaustive cases.^[45] For instance, it resolves borderline cases like \sum_{n=2}^\infty \frac{\ln n}{n^{3/2}} where standard ratios are inconclusive, by selecting p_n such that the modified product limit distinguishes convergence via comparison to \sum 1/n^{3/2}.^[43] Frink's ratio test, formulated by Orrin Frink Jr. in 1948, incorporates an exponential refinement using lim sup of powered ratios: for \sum a_n with a_n >0, let L = \limsup_{n \to \infty} \left( \frac{a_n}{a_{n+1}} \right)^n; the series converges if L < 1/e and diverges if L > 1/e.^[46] This test applies to series like \sum n^{-n}, where the basic ratio L=1 fails, but Frink's L <1/e confirms absolute convergence.^[47] Ali's second ratio test, introduced by Sayel A. Ali in 2008, uses two consecutive ratios to overcome limitations of single-ratio tests. Compute \rho_n = \frac{a_n / a_{n+1}}{a_{n+1} / a_{n+2}}; the series converges if \liminf_{n \to \infty} n (\rho_n - 1) > 1. For \sum_{n=1}^\infty \frac{n}{n^3}, the basic test is inconclusive (L=1), but the second ratio yields lim inf >1, establishing convergence. Ali's m-th ratio test generalizes the second ratio test to m consecutive terms, providing broader applicability. It defines a sequence of ratios over m steps and requires \liminf_{n \to \infty} n \left( \frac{\prod_{k=0}^{m-1} (a_{n+k}/a_{n+k+1})}{\prod_{k=0}^{m-1} (a_{n+1+k}/a_{n+1+k+1})} - 1 \right) > 1 for convergence.^[48] This succeeds for series like \sum_{n=2}^\infty \frac{1}{n^3 \ln n} where m=3 resolves the inconclusive L=1 from basic and second-ratio tests by yielding a limit >1.^[49] Ali-Deutsche Cohen φ-ratio test, developed by Sayel A. Ali and Marion Deutsche Cohen in 2013, introduces a flexible function φ(n) to weight the ratios, enhancing adaptability. The test states that if \liminf_{n \to \infty} \frac{1}{\phi(n)} \ln \left( \frac{a_n}{a_{n+1}} \right) > 1 for a suitable increasing φ (e.g., φ(n)=n), then the series converges. For the series \sum_{n=N}^\infty \frac{1}{n (\ln n)^2 (\ln \ln n)^2} (n large enough for ln ln n defined), basic ratios fail, but choosing appropriate φ yields a lim inf >1, proving convergence.^[50]

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This test, unlike the root test, had already been used by others in special cases. Cauchy proved the ratio test in his Calcul infinitesimal by appealing, as ...
[13]
Ratio and Root Tests - Department of Mathematics at UTSA
Oct 29, 2021 · The test was first published by Jean le Rond d'Alembert and is sometimes known as d'Alembert's ratio test or as the Cauchy ratio test.Missing: De Morgan
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[PDF] Extension of the Bertrand–De Morgan Test and Its Application - arXiv
Apr 9, 2021 · The simplest test for convergence or divergence of se- ries (1) is the ratio test. The first ratio test, the most elementary, was due to d' ...
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5.6 Ratio and Root Tests - Calculus Volume 2 | OpenStax
Mar 30, 2016 · Here we introduce the ratio test, which provides a way of measuring how fast the terms of a series approach zero. Theorem 5.16. Ratio Test. Let ...
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[PDF] Lecture 21: Ratio Test, 10/25/2021
The ratio test was first formulated by Jean d'Alembert. It appears in the work. “Opuscules” published in 1768. Examples. 21.5. Example: Use the ratio test in ...
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[PDF] 11.6 Absolute Convergence and the Ratio and Root Tests
• If L = 1 then the ratio test is inconclusive. Sketch Proof. If L < 1 then r = 1+L. 2 lies half way between L and 1. Taking c = 1−L. 2 in the definition of.
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[PDF] 10.7 Absolute Convergence and the Ratio Test
The Ratio Test has three parts, (a), (b), and (c), and each part requires a separate proof. (a) L<1 " the series is absolutely convergent. The basic pattern of ...
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[PDF] Sequences and Series
THEOREM 11.7.1 The Ratio Test. Suppose that lim n→∞ |an+1/an| = L. If L < 1 the series P an converges absolutely, if L > 1 the series diverges, and if L = 1 ...<|control11|><|separator|>
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[PDF] Lecture 21: The ratio test - Nathan Pflueger
Oct 26, 2011 · Then: • If |L| < 1, then the series converges. If |L| > 1, then the series diverges. Note that if |L| = 1, then the ratio test is inconclusive.Missing: interpretation | Show results with:interpretation
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11.7 The Ratio and Root Tests
The ratio test uses limn→∞|an+1/an| and the root test uses limn→∞|an|1/n. If these limits are <1, the series converges; if >1, it diverges.
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[PDF] Approximating the Sum of a Convergent Series - Larry Riddle
nn n! = n n + 1 n. = 1. 1 + 1 n n which is less than 1 for all n and which decreases to the limit L = 1 e . From inequality (2) we get. (after some ...
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[PDF] Math 115 Exam #1 Practice Problems
Answer: Using the Ratio Test, lim n→∞. (−1)n+1 (n+1)! πn+1. (−1)n n! πn. = lim n→∞ n + 1 π. = ∞. Therefore, the Ratio Test says that the series diverges. 4.<|control11|><|separator|>
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[PDF] Math 131Infinite Series, Part V: The Ratio and Root Tests
Because of the exponential and factorial let's try the ratio test. ... By the ratio test the series converges. EXAMPLE 14.32. Here's a slightly ...
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Calculus II - Integral Test - Pauls Online Math Notes
Nov 16, 2022 · In this section we will discuss using the Integral Test to determine if an infinite series converges or diverges. The Integral Test can be ...
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Calculus II - Alternating Series Test - Pauls Online Math Notes
Nov 16, 2022 · In this section we will discuss using the Alternating Series Test to determine if an infinite series converges or diverges. The Alternating ...
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6.1 Power Series and Functions - Calculus Volume 2 | OpenStax
Mar 30, 2016 · ... power series, we typically apply the ratio test. ... In the following exercises, use the ratio test to determine the radius of convergence of each ...
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Calculus II - Power Series - Pauls Online Math Notes
Nov 16, 2022 · Likewise, if the power series converges for every x the radius of convergence is R=∞ and interval of convergence is −∞<x<∞ − ∞ < x < ∞ .
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[PDF] Power Series - UC Davis Math
The ratio test gives a simple, but useful, way to compute the radius of convergence, although it doesn't apply to every power series. Theorem 6.4. Suppose that ...
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[PDF] 10.4 The Taylor Series for ex,sin x, and cos x -
The ratio test gives convergence if L < 1, which means 1x1 <4. Page 8. 10 Infinite Series x3 x5. EXAMPLE 2 The sine series x - -. + -. - n o - has r = a, (it ...
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[PDF] Bessel Functions - Lecture 7
Thus by the ratio test, the series converges for. 0 <x< ∞. One can use the series to demonstrate the recursion relation between Bessel functions of different ...
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[PDF] Basics of Bessel Functions - PDXScholar
May 26, 2018 · Clearly, this power series is only meaningful if it is convergent. We will check by the well- known Ratio Test. If it passes, we have found ...
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[PDF] Supplement 5: Stirling's Approximation to the Factorial
The ratio n!/s(n) = eεn could also be (and usually is) written as 1 + ηn. Both expressions have the feature that limε→0 eε = limη→0 1 + η = 1.
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[PDF] Necessary and sufficient conditions for the convergence of positive ...
Abstract. We provide new necessary and sufficient conditions for the convergence of positive series developing Bertran–De Morgan and Cauchy type tests given ...
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[PDF] arXiv:2104.01702v1 [math.CA] 4 Apr 2021
Apr 4, 2021 · Bertrand, De Morgan and Kummer. They are classified into the De Morgan hierarchy [3, 4]. The extended Bertrand–De Morgan test is the last test.
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Raabe's Test -- from Wolfram MathWorld
Raabe's test: 1. If rho>1 , the series converges. 2. If rho<1 , the series diverges. 3. If rho=1 , the series may converge or diverge.
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RAABE'S TEST - jstor
Charles Β. Huelsman III, The Ohio State University. Raabe's test, developed by J. L. Raabe in 1832, is a test for the. convergence and divergence of infinite ...
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[PDF] A generalization of Bertrand's test 1. Introduction
One of the most practical routine tests for convergence of a positive series makes use of the ratio test. If this test fails, we can use Rabbe's test. When ...
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Bertrand's Test -- from Wolfram MathWorld
A convergence test also called "de Morgan's and Bertrand's test." If the ratio of terms of a series {a_n}_(n=1)^infty can be written in the form
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[PDF] Advanced Tests for Convergence
May 8, 2019 · These examples include proofs that show convergence or divergence in a variety of ways, including using the Cauchy Criterion for sequences and.
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Kummer's Test Gives Characterizations for Convergence or ... - jstor
Kummer's test gives very powerful sufficient conditions for convergence or divergence of a positive series. As mentioned above it is the source of many other.Missing: modification | Show results with:modification
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[PDF] On Kummer's test of convergence and its relation to basic ... - arXiv
Feb 2, 2018 · In 19th century, Kummer proposed a test of convergence for any positive series based on finding a suitable positive sequence {pn} and a suitable.Missing: modification | Show results with:modification
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Kummer's Test Gives Characterizations for Convergence or ...
Apr 18, 2018 · Kummer's Test Gives Characterizations for Convergence or Divergence of all Positive Series: The American Mathematical Monthly: Vol 101, No 5.Missing: modification | Show results with:modification
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tive in some cases where the d'Alembert test fail
A series of positive terms (1) ^an converges if lining (an/an-i)n<l/e, and diverges if limn+<x>(an/an-i)n>l/e. More generally, (1) converges if lim sup (an/an-i) ...
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[PDF] On a ratio test of Frink - ACDSee 32 print job - ICM
For k-1 it follows that the series satisfying Frink's test of convergence are iden- tical with those satisfying the test of Raabe. Indeed, suppose (1). Then.
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A Second Look at the Second Ratio Test - Taylor & Francis Online
This article examines a class of series convergence tests, known as the mth ratio tests, that were introduced by Sayel A. Ali in 2008.
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[PDF] A Second Raabe's Test and Other Series Tests - arXiv
Sep 12, 2021 · Le Rond D'Alembert formulated D'Alembert's Ratio Test, otherwise known today as the Ratio Test. D'Alembert felt that the theory of limits needed.
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The phi-ratio tests - EMS Press
The phi-ratio tests. Sayel Ali and Marion Deutsche Cohen. Sayel Ali received his B.Sc. in Mathematics from the University of Jordan, M.Sc. in. Mathematics from ...