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Ratio test
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=== De Morgan hierarchy === [[Augustus De Morgan]] proposed a hierarchy of ratio-type tests<ref name="Bromwich1908"/><ref name="Blackburn2012"/> The ratio test parameters (<math>\rho_n</math>) below all generally involve terms of the form <math>D_n a_n/a_{n+1}-D_{n+1}</math>. This term may be multiplied by <math>a_{n+1}/a_n</math> to yield <math>D_n-D_{n+1}a_{n+1}/a_n</math>. This term can replace the former term in the definition of the test parameters and the conclusions drawn will remain the same. Accordingly, there will be no distinction drawn between references which use one or the other form of the test parameter. ====1. d'Alembert's ratio test==== The first test in the De Morgan hierarchy is the ratio test as described above. ====2. Raabe's test==== This extension is due to [[Joseph Ludwig Raabe]]. Define: :<math>\rho_n \equiv n\left(\frac{a_n}{a_{n+1}}-1\right)</math> (and some extra terms, see Ali, Blackburn, Feld, Duris (none), Duris2){{Clarify|date=September 2024}} The series will:<ref name="Ali2008"/><ref name="Duris2009"/><ref name="Blackburn2012"/> * Converge when there exists a ''c>''1 such that <math>\rho_n \ge c</math> for all ''n>N''. * Diverge when <math>\rho_n \le 1</math> for all ''n>N''. * Otherwise, the test is inconclusive. For the limit version,<ref>{{mathworld|title=Raabe's Test|urlname=RaabesTest}}</ref> the series will: * Converge if <math>\rho=\lim_{n\to\infty}\rho_n>1</math> (this includes the case ''ρ'' = ∞) * Diverge if <math>\lim_{n\to\infty}\rho_n<1</math>. * If ''ρ'' = 1, the test is inconclusive. When the above limit does not exist, it may be possible to use limits superior and inferior.<ref name="Bromwich1908"/> The series will: * Converge if <math>\liminf_{n \to \infty} \rho_n > 1</math> * Diverge if <math>\limsup_{n \rightarrow \infty} \rho_n < 1</math> * Otherwise, the test is inconclusive. =====Proof of Raabe's test===== Defining <math>\rho_n \equiv n\left(\frac{a_n}{a_{n+1}}-1\right)</math>, we need not assume the limit exists; if <math>\limsup\rho_n<1</math>, then <math>\sum a_n</math> diverges, while if <math>\liminf \rho_n>1</math> the sum converges. The proof proceeds essentially by comparison with <math>\sum1/n^R</math>. Suppose first that <math>\limsup\rho_n<1</math>. Of course if <math>\limsup\rho_n<0</math> then <math>a_{n+1}\ge a_n</math> for large <math>n</math>, so the sum diverges; assume then that <math>0\le\limsup\rho_n<1</math>. There exists <math>R<1</math> such that <math>\rho_n\le R</math> for all <math>n\ge N</math>, which is to say that <math>a_{n}/a_{n+1}\le \left(1+\frac Rn\right)\le e^{R/n}</math>. Thus <math>a_{n+1}\ge a_ne^{-R/n}</math>, which implies that <math>a_{n+1}\ge a_Ne^{-R(1/N+\dots+1/n)}\ge ca_Ne^{-R\log(n)}=ca_N/n^R</math> for <math>n\ge N</math>; since <math>R<1</math> this shows that <math>\sum a_n</math> diverges. The proof of the other half is entirely analogous, with most of the inequalities simply reversed. We need a preliminary inequality to use in place of the simple <math>1+t<e^t</math> that was used above: Fix <math>R</math> and <math>N</math>. Note that <math>\log\left(1+\frac Rn\right)=\frac Rn+O\left(\frac 1{n^2}\right)</math>. So <math>\log\left(\left(1+\frac RN\right)\dots\left(1+\frac Rn\right)\right) =R\left(\frac 1N+\dots+\frac 1n\right)+O(1)=R\log(n)+O(1)</math>; hence <math>\left(1+\frac RN\right)\dots\left(1+\frac Rn\right)\ge cn^R</math>. Suppose now that <math>\liminf\rho_n>1</math>. Arguing as in the first paragraph, using the inequality established in the previous paragraph, we see that there exists <math>R>1</math> such that <math>a_{n+1}\le ca_Nn^{-R}</math> for <math>n\ge N</math>; since <math>R>1</math> this shows that <math>\sum a_n</math> converges. ==== 3. Bertrand's test ==== This extension is due to [[Joseph Bertrand]] and [[Augustus De Morgan]]. Defining: :<math>\rho_n \equiv n \ln n\left(\frac{a_n}{a_{n+1}}-1\right)-\ln n</math> Bertrand's test<ref name="Bromwich1908"/><ref name="Duris2009"/> asserts that the series will: * Converge when there exists a ''c>1'' such that <math>\rho_n \ge c</math> for all ''n>N''. * Diverge when <math>\rho_n \le 1</math> for all ''n>N''. * Otherwise, the test is inconclusive. For the limit version, the series will: * Converge if <math>\rho=\lim_{n\to\infty}\rho_n>1</math> (this includes the case ''ρ'' = ∞) * Diverge if <math>\lim_{n\to\infty}\rho_n<1</math>. * If ''ρ'' = 1, the test is inconclusive. When the above limit does not exist, it may be possible to use limits superior and inferior.<ref name="Bromwich1908"/><ref name="Blackburn2012"/><ref>{{mathworld|title=Bertrand's Test|urlname=BertrandsTest}}</ref> The series will: * Converge if <math>\liminf \rho_n > 1</math> * Diverge if <math>\limsup \rho_n < 1</math> * Otherwise, the test is inconclusive. ==== 4. Extended Bertrand's test ==== This extension probably appeared at the first time by Margaret Martin in 1941.<ref name="Mar1941">{{cite journal|url=https://www.ams.org/journals/bull/1941-47-06/S0002-9904-1941-07477-X/S0002-9904-1941-07477-X.pdf |last1=Martin |first1=Margaret |date=1941 |title=A sequence of limit tests for the convergence of series |journal=Bulletin of the American Mathematical Society |volume=47 |issue=6|pages=452–457 |doi=10.1090/S0002-9904-1941-07477-X |doi-access=free }}</ref> A short proof based on Kummer's test and without technical assumptions (such as existence of the limits, for example) was provided by Vyacheslav Abramov in 2019.<ref name="Abr2008">{{cite journal|last1=Abramov |first1=Vyacheslav M. |date=May 2020 |title=Extension of the Bertrand–De Morgan test and its application |journal=The American Mathematical Monthly |volume=127 |issue=5 |pages=444–448 |doi=10.1080/00029890.2020.1722551 |arxiv=1901.05843 |s2cid=199552015 }}</ref> Let <math>K\geq1</math> be an integer, and let <math>\ln_{(K)}(x)</math> denote the <math>K</math>th [[iteration|iterate]] of [[natural logarithm]], i.e. <math>\ln_{(1)}(x)=\ln (x)</math> and for any <math>2\leq k\leq K</math>, <math>\ln_{(k)}(x)=\ln_{(k-1)}(\ln (x))</math>. Suppose that the ratio <math>a_n/a_{n+1}</math>, when <math>n</math> is large, can be presented in the form :<math>\frac{a_n}{a_{n+1}}=1+\frac{1}{n}+\frac{1}{n}\sum_{i=1}^{K-1}\frac{1}{\prod_{k=1}^i\ln_{(k)}(n)}+\frac{\rho_n}{n\prod_{k=1}^K\ln_{(k)}(n)}, \quad K\geq1.</math> (The empty sum is assumed to be 0. With <math>K=1</math>, the test reduces to Bertrand's test.) The value <math>\rho_{n}</math> can be presented explicitly in the form :<math>\rho_{n} = n\prod_{k=1}^K\ln_{(k)}(n)\left(\frac{a_n}{a_{n+1}}-1\right)-\sum_{j=1}^K\prod_{k=1}^j\ln_{(K-k+1)}(n).</math> Extended Bertrand's test asserts that the series * Converge when there exists a <math>c>1</math> such that <math>\rho_n \geq c</math> for all <math>n>N</math>. * Diverge when <math>\rho_n \leq 1</math> for all <math>n>N</math>. * Otherwise, the test is inconclusive. For the limit version, the series * Converge if <math>\rho=\lim_{n\to\infty}\rho_n>1</math> (this includes the case <math>\rho = \infty</math>) * Diverge if <math>\lim_{n\to\infty}\rho_n<1</math>. * If <math>\rho = 1</math>, the test is inconclusive. When the above limit does not exist, it may be possible to use limits superior and inferior. The series * Converge if <math>\liminf \rho_n > 1</math> * Diverge if <math>\limsup \rho_n < 1</math> * Otherwise, the test is inconclusive. For applications of Extended Bertrand's test see [[birth–death process]]. ==== 5. Gauss's test ==== This extension is due to [[Carl Friedrich Gauss]]. Assuming ''a<sub>n</sub>'' > 0 and ''r > 1'', if a bounded sequence ''C<sub>n</sub>'' can be found such that for all ''n'':<ref name="Knopp"/><ref name="Ali2008"/><ref name="Blackburn2012"/><ref name="Duris2009"/> :<math>\frac{a_n}{a_{n+1}}=1+\frac{\rho}{n}+\frac{C_n}{n^r}</math> then the series will: * Converge if <math>\rho>1</math> * Diverge if <math>\rho \le 1</math> ==== 6. Kummer's test ==== This extension is due to [[Ernst Kummer]]. Let ζ<sub>''n''</sub> be an auxiliary sequence of positive constants. Define :<math>\rho_n \equiv \left(\zeta_n \frac{a_n}{a_{n+1}} - \zeta_{n+1}\right)</math> Kummer's test states that the series will:<ref name="Knopp"/><ref name="Tong1994"/><ref name="Duris2009"/><ref name="Duris2018"/> * Converge if there exists a <math>c>0</math> such that <math>\rho_n \ge c</math> for all n>N. (Note this is not the same as saying <math>\rho_n > 0</math>) * Diverge if <math>\rho_n \le 0</math> for all n>N and <math>\sum_{n=1}^\infty 1/\zeta_n</math> diverges. For the limit version, the series will:<ref>{{mathworld|title=Kummer's Test|urlname=KummersTest}}</ref><ref name="Ali2008"/><ref name="Blackburn2012"/> * Converge if <math>\lim_{n\to\infty}\rho_n>0</math> (this includes the case ''ρ'' = ∞) * Diverge if <math>\lim_{n\to\infty}\rho_n<0</math> and <math>\sum_{n=1}^\infty 1/\zeta_n</math> diverges. * Otherwise the test is inconclusive When the above limit does not exist, it may be possible to use limits superior and inferior.<ref name="Bromwich1908"/> The series will * Converge if <math>\liminf_{n \to \infty} \rho_n >0</math> * Diverge if <math>\limsup_{n \to \infty} \rho_n <0</math> and <math>\sum 1/\zeta_n</math> diverges. ===== Special cases ===== All of the tests in De Morgan's hierarchy except Gauss's test can easily be seen as special cases of Kummer's test:<ref name="Bromwich1908"/> * For the ratio test, let ζ<sub>n</sub>=1. Then: ::<math>\rho_\text{Kummer} = \left(\frac{a_n}{a_{n+1}}-1\right) = 1/\rho_\text{Ratio}-1</math> * For Raabe's test, let ζ<sub>n</sub>=n. Then: ::<math>\rho_\text{Kummer} = \left(n\frac{a_n}{a_{n+1}}-(n+1)\right) = \rho_\text{Raabe}-1</math> * For Bertrand's test, let ζ<sub>n</sub>=n ln(n). Then: ::<math>\rho_\text{Kummer} = n \ln(n)\left(\frac{a_n}{a_{n+1}}\right)-(n+1)\ln(n+1)</math> :Using <math>\ln(n+1)=\ln(n)+\ln(1+1/n)</math> and [[approximation|approximating]] <math>\ln(1+1/n)\rightarrow 1/n</math> for large ''n'', which is negligible compared to the other terms, <math>\rho_\text{Kummer}</math> may be written: ::<math>\rho_\text{Kummer} = n \ln(n)\left(\frac{a_n}{a_{n+1}}-1\right)-\ln(n)-1 = \rho_\text{Bertrand}-1</math> * For Extended Bertrand's test, let <math>\zeta_n=n\prod_{k=1}^K\ln_{(k)}(n).</math> From the [[Taylor series]] expansion for large <math>n</math> we arrive at the [[approximation]] ::<math>\ln_{(k)}(n+1)=\ln_{(k)}(n)+\frac{1}{n\prod_{j=1}^{k-1}\ln_{(j)}(n)}+O\left(\frac{1}{n^2}\right),</math> where the empty product is assumed to be 1. Then, ::<math>\rho_\text{Kummer} = n\prod_{k=1}^K\ln_{(k)}(n)\frac{a_n}{a_{n+1}}-(n+1)\left[\prod_{k=1}^K\left(\ln_{(k)}(n)+\frac{1}{n\prod_{j=1}^{k-1}\ln_{(j)}(n)}\right)\right]+o(1) =n\prod_{k=1}^K\ln_{(k)}(n)\left(\frac{a_n}{a_{n+1}}-1\right)-\sum_{j=1}^K\prod_{k=1}^j\ln_{(K-k+1)}(n)-1+o(1).</math> Hence, ::<math>\rho_\text{Kummer} = \rho_\text{Extended Bertrand}-1.</math> Note that for these four tests, the higher they are in the De Morgan hierarchy, the more slowly the <math>1/\zeta_n</math> series diverges. =====Proof of Kummer's test===== If <math>\rho_n>0</math> then fix a positive number <math>0<\delta<\rho_n</math>. There exists a natural number <math>N</math> such that for every <math>n>N,</math> :<math>\delta\leq\zeta_{n}\frac{a_{n}}{a_{n+1}}-\zeta_{n+1}.</math> Since <math>a_{n+1}>0</math>, for every <math>n> N,</math> :<math>0\leq \delta a_{n+1}\leq \zeta_{n}a_{n}-\zeta_{n+1}a_{n+1}.</math> In particular <math>\zeta_{n+1}a_{n+1}\leq \zeta_{n}a_{n}</math> for all <math>n\geq N</math> which means that starting from the index <math>N</math> the sequence <math>\zeta_{n}a_{n}>0</math> is monotonically decreasing and positive which in particular implies that it is bounded below by 0. Therefore, the limit :<math>\lim_{n\to\infty}\zeta_{n}a_{n}=L</math> exists. This implies that the positive [[telescoping series]] :<math>\sum_{n=1}^{\infty}\left(\zeta_{n}a_{n}-\zeta_{n+1}a_{n+1}\right)</math> is convergent, and since for all <math>n>N,</math> :<math>\delta a_{n+1}\leq \zeta_{n}a_{n}-\zeta_{n+1}a_{n+1}</math> by the [[direct comparison test]] for positive series, the series <math>\sum_{n=1}^{\infty}\delta a_{n+1}</math> is convergent. On the other hand, if <math>\rho<0</math>, then there is an ''N'' such that <math>\zeta_n a_n</math> is increasing for <math>n>N</math>. In particular, there exists an <math>\epsilon>0</math> for which <math>\zeta_n a_n>\epsilon</math> for all <math>n>N</math>, and so <math>\sum_n a_n=\sum_n \frac{a_n\zeta_n}{\zeta_n}</math> diverges by comparison with <math>\sum_n \frac \epsilon {\zeta_n}</math>.
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