|
|
| Line 1: |
Line 1: |
| {{Calculus|Series}}
| | Anyone invest loads of money on things like controls or alternatively memory cards, appear the net for a secondhand variations. Occasionally a store will probably get out of used-game hardware, which could be inexpensive. If you loved this article and you would such as to receive more details pertaining to [http://circuspartypanama.com Clash of Clans hack no survey] kindly visit our own internet site. Make sure you look for any web-based seller's feedback prior to making the purchase so this whether you are procuring what you [http://Www.Reddit.com/r/howto/search?q=covered covered].<br><br>Shelter are able to turn to those gems to [http://en.search.Wordpress.com/?q=instantly+fortify instantly fortify] his army. He tapped 'Yes,'" close to without thinking. Back under a month to do with walking around a variety of hours on a daily basis, he''d spent near 1000 dollars.<br><br>Prevent purchasing big title events near their launch dates. Waiting means that you're prone to have clash of clans cheats after having a patch or two has recently emerge to mend obvious holes and bugs that may impact your pleasure also game play. At the same time keep an eye out in the open for titles from broadcasters which are understood our bodies and wellbeing patching and support.<br><br>Games consoles game playing is ideal for kids. Consoles will offer you far better control pertaining to content and safety, as often kids can simply wind turbine by way of elder regulates on your computer. Using this step might help defend your young ones provided by harm.<br><br>You should may not let computer games take over your days. Game titles can be quite additive, time also have to make indeed you moderate the period of time that you investing playing such games. If you invest an excessive level of time playing video game, your actual life will begin to falter.<br><br>That tutorial will guide you through your first few raids, constructions, and upgrades, while youre left to private wiles pretty quickly. Your buildings take real time to construct and upgrade, your army units take time to recruit, and your bio buildings take time to get food and gold. Like all of its genre cousins, Throne Haste is meant to took part in multiple short bursts in the daytlight. This type of compulsive gaming definitely works more beneficial on mobile devices usually are always with you and could send push notifications when timed tasks are completed. Then again, the success of a lot of hit Facebook games through the years indicates that people consult Facebook often enough different short play sessions employment there too.<br><br>Plainly individuals who produced this unique Crack Clash of Family members are true fans having to do with the sport themselves, not to mention this is exactly the things ensures the potency in our alternative, because my wife and i needed to do it ourselves. |
| | |
| In [[mathematics]], the '''ratio test''' is a [[convergence tests|test]] (or "criterion") for the [[convergent series|convergence]] of a [[series (mathematics)|series]] <math>\sum_{n=1}^\infty a_n</math>, where each term is a real or complex number and <math>a_n</math> is nonzero when ''n'' is large. 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'''.<ref name=":0">http://mathworld.wolfram.com/RatioTest.html</ref>
| |
| | |
| == Motivation ==
| |
| | |
| Given the following [[geometric series]]:
| |
| :<math>\sum_{n=1}^\infty \left(\frac{1}{2}\right)^n = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots</math>
| |
| The quotient <math>a_{n+1}/a_n = (1/2)^{n+1}/(1/2)^n</math> of any two adjacent terms is 1/2. The sum of the first ''m'' terms is given by:
| |
| :<math>1 - \frac{1}{2^m}.</math>
| |
| As ''m'' increases, this converges to 1, so the sum of the series is 1. On the other hand given this geometric series:
| |
| :<math>\sum_{n=1}^\infty 2^n = 2 + 4 + 8 + \cdots</math>
| |
| The quotient <math>a_{n+1}/a_n</math> of any two adjacent terms is 2. The sum of the first ''m'' terms is given by
| |
| :<math>2^{m+1} - 2,</math>
| |
| which increases without bound as ''m'' increases, so this series diverges. More generally, the [[Geometric_progression#Geometric_series|sum of the first ''m'' terms of the geometric series]] <math>\sum_{n=1}^\infty r^n</math> is given by:
| |
| :<math>\sum_{n=1}^{m} r^n = \frac{r}{r-1} (r^m - 1).</math>
| |
| Whether this converges or diverges as ''m'' increases depends on whether ''r'', the quotient of any two adjacent terms, is less than or greater than 1. Now consider the series:
| |
| :<math>\sum_{n=1}^\infty \frac{n+1}{n} \left(\frac{1}{2}\right)^n = \frac{2}{1} \cdot \frac{1}{2} + \frac{3}{2} \cdot \frac{1}{4} + \frac{4}{3} \cdot \frac{1}{8} + \cdots</math>
| |
| This is similar to the first convergent sequence above, except that now the ratio of two terms is not fixed at exactly 1/2:
| |
| :<math>\left(\frac{n+1}{n} \left(\frac{1}{2}\right)^n\right)/\left(\frac{n}{n-1} \left(\frac{1}{2}\right)^{n-1}\right) = \frac{n^2-1}{2n^2} = \frac{1}{2} - \frac{1}{2n^2}.</math>
| |
| However, as ''n'' increases, the ratio still tends in the limit towards the same constant 1/2. The ratio test generalizes the simple test for geometric series to more complex series like this one where the quotient of two terms is not fixed, but in the limit tends towards a fixed value. The rules are similar: if the quotient approaches a value less than one, the series converges, whereas if it approaches a value greater than one, the series diverges.
| |
| | |
| == The test ==
| |
| The usual form of the test makes use of the [[limit (mathematics)|limit]]
| |
| {{NumBlk|:|<math>L = \lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|.</math>|{{EquationRef|1}}}}
| |
| The ratio test states that:
| |
| * if ''L'' < 1 then the series [[absolute convergence|converges absolutely]];
| |
| * if ''L'' > 1 then the series does not converge;
| |
| * if ''L'' = 1 or the limit fails to exist, then the test is inconclusive, because there exist both convergent and divergent series that satisfy this case.
| |
| | |
| It is possible to make the ratio test applicable to certain cases where the limit ''L'' fails to exist, if [[limit superior]] and [[limit inferior]] are used. The test criteria can also be refined so that the test is sometimes conclusive even when ''L'' = 1. More specifically, let
| |
| :<math>R = \lim\sup \left|\frac{a_{n+1}}{a_n}\right|</math> and <math>r = \lim\inf \left|\frac{a_{n+1}}{a_n}\right|</math>. | |
| Then the ratio test states that:<ref>{{harvnb|Rudin|1976|loc=§3.34}}</ref><ref>{{harvnb|Apostol|1974|loc=§8.14}}</ref>
| |
| * if ''R'' < 1, the series converges absolutely;
| |
| * if ''r'' > 1, the series diverges;
| |
| * if <math>\left|\frac{a_{n+1}}{a_n}\right|\ge 1</math> for all large ''n'' (regardless of the value of ''r'' ), the series also diverges; this is because <math>|a_n|</math> is nonzero and increasing and hence <math>a_n</math> does not approach zero;
| |
| * the test is otherwise inconclusive.
| |
| | |
| If the limit ''L'' in ({{EquationNote|1}}) exists, we must have ''L=R=r''. So the original ratio test is a weaker version of the refined one.
| |
| | |
| == Examples ==
| |
| | |
| === Convergent because ''L''<1 ===
| |
| Consider the series
| |
| | |
| :<math>\sum_{n=1}^\infty\frac{n}{e^n}</math>
| |
| | |
| Putting this into the ratio test:
| |
| | |
| :<math>L = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|
| |
| = \lim_{n\to\infty} \left| \frac{\frac{n+1}{e^{n+1}}}{\frac{n}{e^n}}\right|
| |
| = \frac{1}{e} < 1.</math>
| |
| | |
| Thus the series converges.
| |
| | |
| === Divergent because ''L>1'' ===
| |
| Consider the series
| |
| | |
| :<math>\sum_{n=1}^\infty\frac{e^n}{n}.</math>
| |
| | |
| Putting this into the ratio test:
| |
| | |
| :<math>L
| |
| = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|
| |
| = \lim_{n\to\infty} \left| \frac{\frac{e^{n+1}}{n+1}}{\frac{e^n}{n}} \right|
| |
| = e > 1.</math>
| |
| | |
| Thus the series diverges.
| |
| | |
| === Inconclusive because ''L''=1 ===
| |
| Consider the three series
| |
| :<math>\sum_{n=1}^\infty 1,</math> <math>\sum_{n=1}^\infty \frac{1}{n^2}</math> and <math>\sum_{n=1}^\infty (-1)^n\frac{1}{n}</math>.
| |
| The first series diverges, the second one converges absolutely and the third one converges conditionally. However, the term-by-term magnitude ratios <math>\left|\frac{a_{n+1}}{a_n}\right|</math> of the three series are respectively 1, <math>\frac{n^2}{(n+1)^2}</math> and <math>\frac{n}{n+1}</math>. So, in all three cases, we have<math>\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|=1</math>. This illustrates that when ''L''=1, the series may converge or diverge and hence the original ratio test is inconclusive. For the first series <math>\sum_{n=1}^\infty 1</math>, however, as the term-by-term magnitude ratio <math>\left|\frac{a_{n+1}}{a_n}\right|=1</math> for all ''n'', we can apply the third criterion in the refined version of the ratio test to conclude that the series diverges.
| |
| | |
| == Proof ==
| |
| [[File:Ratio test proof.svg|thumb|300px|In this example, the ratio of adjacent terms in the blue sequence converges to L=1/2. We choose ''r'' = (L+1)/2 = 3/4. Then the blue sequence is dominated by the red sequence ''r''<sup>''k''</sup> for all ''n'' ≥ 2. The red sequence converges, so the blue sequence does as well.]]
| |
| Below is a proof of the validity of the original ratio test.
| |
| | |
| Suppose that <math>L = \lim_{n\rightarrow\infty} \left| \frac{a_{n+1}}{a_{n}}\right| < 1</math>. We can then show that the series converges absolutely by showing that its terms will eventually become less than those of a certain convergent [[geometric series]]. To do this, let <math>r = \frac{L+1}{2}</math>. Then ''r'' is strictly between ''L'' and 1, and <math>|a_{n+1}| < r |a_{n}|</math> for sufficiently large n (say, n greater than N). Hence <math>|a_{n+i}| < r^i|a_{n}|</math> for each ''n > N'' and ''i > 0'', and so
| |
| | |
| :<math>\sum_{i=N+1}^{\infty}|a_{i}| = \sum_{i=1}^{\infty}|a_{N+i}|
| |
| < \sum_{i=1}^{\infty}r^{i}|a_{N+1}| = |a_{N+1}|\sum_{i=1}^{\infty}r^{i}
| |
| = |a_{N+1}|\frac{r}{1 - r} < \infty.</math>
| |
| | |
| That is, the series converges absolutely.
| |
| | |
| On the other hand, if ''L'' > 1, then <math>|a_{n+1}| > |a_{n}|</math> for sufficiently large n, so that the limit of the summands is non-zero. Hence the series diverges.
| |
| | |
| == Extensions for ''L''=1 ==
| |
| {{Expand section|De Morgan's hierarchy and usage examples.
| |
| And proofs should be provided. |date=August 2013}}
| |
| | |
| As seen in the previous example, the ratio test may be inconclusive when the limit of the ratio is 1. Extensions to ratio test, however, sometimes allows one to deal with this case. For instance, the aforementioned refined version of the test handles the case <math>\left|\frac{a_{n+1}}{a_n}\right|\ge1</math>. Below are some other extensions.
| |
| | |
| ===Raabe's test===
| |
| This extension is due to [[Joseph Ludwig Raabe]]. It states that if
| |
| | |
| :<math>\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|=1</math>
| |
| | |
| :<math>\lim_{n\rightarrow\infty}
| |
| \,n\left(\,\left|\frac{a_{n+1}}{a_n}\right|-1\right)<-1</math>
| |
| | |
| then the series will be absolutely convergent.<ref>http://mathworld.wolfram.com/RaabesTest.html</ref> d'Alembert's ratio test and Raabe's test are the first and second theorem in a hierarchy of such theorems due to [[Augustus De Morgan]].{{Citation needed|reason = The existence/content of De Morgan's hierarchy.|date = August 2013}}
| |
| | |
| ===Higher order tests===
| |
| The next cases in de Morgan's hierarchy are Bertrand's and Gauss's test. Each test involves slightly different higher order asymptotics. If
| |
| :<math>\left|\frac{a_n}{a_{n+1}}\right| = 1 + \frac{1}{n} + \frac{\rho_n}{n\ln n}</math>
| |
| then the series converges if lim inf ρ<sub>''n''</sub> > 1, and diverges if lim sup ρ<sub>''n''</sub> < 1. This is '''Bertrand's test'''.<ref>http://mathworld.wolfram.com/BertrandsTest.html</ref>
| |
| | |
| If
| |
| :<math>\left|\frac{a_n}{a_{n+1}}\right| = 1+ \frac{h}{n} + \frac{C_n}{n^r}</math>
| |
| where ''r'' > 1 and ''C''<sub>''n''</sub> is bounded, then the series converges if ''h'' > 1 and diverges if ''h'' ≤ 1. This is '''Gauss's test'''.<ref>http://mathworld.wolfram.com/GausssTest.html</ref>
| |
| | |
| These are both special cases of '''Kummer's test''' for the convergence of the series Σ''a''<sub>''n''</sub>. Let ζ<sub>''n''</sub> be an auxiliary sequence of positive constants. Let
| |
| :<math>\rho = \lim_{n\to\infty} \left(\zeta_n \frac{a_n}{a_{n+1}} - \zeta_{n+1}\right).</math>
| |
| Then if ''ρ'' > 0, the series converges. If ''ρ'' < 0 and Σ1/''ζ''<sub>''n''</sub> diverges, then the series diverges. Otherwise the test is inconclusive.<ref>http://mathworld.wolfram.com/KummersTest.html</ref>
| |
| | |
| ==See also==
| |
| * [[Root test]]
| |
| * [[Radius of convergence]]
| |
| | |
| == Footnotes ==
| |
| {{reflist}}
| |
| | |
| == References ==
| |
| | |
| *{{citation | last1=d'Alembert| first1=J.|authorlink=Jean le Rond d'Alembert| year=1768 | title=Opuscules |volume=V|pages=171–183| url=http://gallica.bnf.fr/ark:/12148/bpt6k62424s.image.f192}}.
| |
| | |
| *{{Citation | last1=Apostol | first1=Tom M. | author1-link=Tom M. Apostol | title=Mathematical analysis | publisher=[[Addison-Wesley]] | edition=2nd | isbn=978-0-201-00288-1 | year=1974}}: §8.14.
| |
| | |
| * {{citation|last=Knopp|first=Konrad|title=Infinite Sequences and Series|publisher=Dover publications, Inc.|publication-place=New York|year=1956|isbn=0-486-60153-6}}: §3.3, 5.4.
| |
| | |
| * {{citation|last=Rudin|first=Walter|authorlink=Walter Rudin|title=Principles of Mathematical Analysis|edition=3rd|publisher=McGraw-Hill, Inc.|publication-place=New York|year=1976|isbn=0-07-054235-X}}: §3.34.
| |
| | |
| * {{springer|title=Bertrand criterion|id=p/b015780}}
| |
| * {{springer|title=Gauss criterion|id=p/g043420}}
| |
| * {{springer|title=Kummer criterion|id=p/k055950}}
| |
| * [http://www.encyclopediaofmath.org/index.php/D'Alembert_criterion D'Alembert criterion] at [http://www.encyclopediaofmath.org/ Encyclopedia of Mathematics]
| |
| * [http://www.encyclopediaofmath.org/index.php/Raabe_criterion Raabe criterion] at [http://www.encyclopediaofmath.org/ Encyclopedia of Mathematics]
| |
| | |
| * {{mathworld|title=Bertrand's Test|urlname=BertrandsTest}}
| |
| | |
| * {{mathworld|title=Gauss's Test|urlname=GausssTest}}
| |
| | |
| * {{mathworld|title=Kummer's Test|urlname=KummersTest}}
| |
| | |
| * {{citation|last2=Whittaker|first2=E. T.|last1=Watson|first1=G. N.|title=A Course in Modern Analysis|edition=4th|publisher=Cambridge University Press|year=1963|isbn=0-521-58807-3}}: §2.36, 2.37.
| |
| | |
| [[Category:Convergence tests]]
| |
| [[Category:Articles containing proofs]]
| |
| | |
| [[it:Criteri di convergenza#Criterio del rapporto (o di d'Alembert)]]
| |
Anyone invest loads of money on things like controls or alternatively memory cards, appear the net for a secondhand variations. Occasionally a store will probably get out of used-game hardware, which could be inexpensive. If you loved this article and you would such as to receive more details pertaining to Clash of Clans hack no survey kindly visit our own internet site. Make sure you look for any web-based seller's feedback prior to making the purchase so this whether you are procuring what you covered.
Shelter are able to turn to those gems to instantly fortify his army. He tapped 'Yes,'" close to without thinking. Back under a month to do with walking around a variety of hours on a daily basis, hed spent near 1000 dollars.
Prevent purchasing big title events near their launch dates. Waiting means that you're prone to have clash of clans cheats after having a patch or two has recently emerge to mend obvious holes and bugs that may impact your pleasure also game play. At the same time keep an eye out in the open for titles from broadcasters which are understood our bodies and wellbeing patching and support.
Games consoles game playing is ideal for kids. Consoles will offer you far better control pertaining to content and safety, as often kids can simply wind turbine by way of elder regulates on your computer. Using this step might help defend your young ones provided by harm.
You should may not let computer games take over your days. Game titles can be quite additive, time also have to make indeed you moderate the period of time that you investing playing such games. If you invest an excessive level of time playing video game, your actual life will begin to falter.
That tutorial will guide you through your first few raids, constructions, and upgrades, while youre left to private wiles pretty quickly. Your buildings take real time to construct and upgrade, your army units take time to recruit, and your bio buildings take time to get food and gold. Like all of its genre cousins, Throne Haste is meant to took part in multiple short bursts in the daytlight. This type of compulsive gaming definitely works more beneficial on mobile devices usually are always with you and could send push notifications when timed tasks are completed. Then again, the success of a lot of hit Facebook games through the years indicates that people consult Facebook often enough different short play sessions employment there too.
Plainly individuals who produced this unique Crack Clash of Family members are true fans having to do with the sport themselves, not to mention this is exactly the things ensures the potency in our alternative, because my wife and i needed to do it ourselves.