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| {{distinguish|Cauchy's convergence test}}
| | Hello. Let me introduce the author. Her title is Emilia Shroyer but it's not the most female title out there. What I love performing is taking part in baseball but I haven't produced a dime with it. He used to be unemployed but now he is a meter reader. Her spouse and her reside in Puerto Rico but she will have to transfer 1 day or an additional.<br><br>Feel free to visit my web page - [http://kard.dk/?p=17207 kard.dk] |
| {{Calculus |Series}}
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| In [[mathematics]], the '''Cauchy condensation test''', named after [[Augustin-Louis Cauchy]], is a standard [[convergence test]] for [[infinite series]]. For a positive [[non-increasing]] [[sequence]] ''f''(''n''), the sum
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| :<math>\sum_{n=1}^{\infty}f(n)</math>
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| converges if and only if the sum
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| :<math>\sum_{n=0}^{\infty} 2^{n}f(2^{n})</math>
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| converges. Moreover, in that case we have
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| :<math>\sum_{n=1}^{\infty}f(n) \leq \sum_{n=0}^{\infty} 2^{n}f(2^{n}) \leq 2 \sum_{n=1}^{\infty}f(n). </math>
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| A geometric view is that we are approximating the sum with [[trapezoid]]s at every <math>2^{n}</math>. Another explanation is that, as with the analogy between finite sums and [[integral]]s, the 'condensation' of terms is analogous to a substitution of an exponential function. This becomes clearer in examples such as
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| :<math>\ f(n) = n^{-a} (\log n)^{-b} (\log \log n)^{-c}.</math>
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| Here the series definitely converges for ''a'' > 1, and diverges for ''a'' < 1. When ''a'' = 1, the condensation transformation essentially gives the series
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| :<math>\sum n^{-b} (\log n)^{-c}.</math>
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| The logarithms 'shift to the left'. So when ''a'' = 1, we have convergence for ''b'' > 1, divergence for ''b'' < 1. When ''b'' = 1 the value of ''c'' enters.
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| == Proof ==
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| Let ''f''(''n'') be a positive, non-increasing sequence of real numbers. To simplify the notation, we will write ''a''<sub>''n''</sub> = ''f''(''n''). We are to investigate the series <math>a_1+a_2+a_3+\cdots</math>. The condensation test follows from noting that if we collect the terms of the series into groups of lengths <math>2^{n}</math>, each of these groups will be less than <math>2^{n} a_{2^{n}}</math> by monotonicity. Observe,
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| :<math>\begin{align}
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| \sum_{n=1}^{\infty} a_n & = a_1+\underbrace{a_2+a_3}_{\leq a_2+a_2}+\underbrace{a_4+a_5+a_6+a_7}_{\leq a_4+a_4+a_4+a_4}+\cdots +\underbrace{a_{2^n}+a_{2^n+1}+\cdots +a_{2^{n+1}-1}}_{\leq a_{2^n}+a_{2^n}+\cdots +a_{2^n}}+\cdots \\
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| & \leq a_1 + 2 a_2 + 4 a_4 + \cdots + 2^n a_{2^n} + \cdots = \sum_{n=0}^{\infty} 2^n a_{2^n}.
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| \end{align}</math>
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| We have used the fact that the sequence ''a''<sub>''n''</sub> is non-increasing, thus <math>a_n\leq a_m</math> whenever <math>n\geq m</math>. The convergence of the original series now follows from direct comparison to this "condensed" series. To see that convergence of the original series implies the convergence of this last series, we similarly put,
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| :<math>\begin{align}
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| \sum_{n=0}^{\infty} 2^n a_{2^n} & = \underbrace{a_1+a_2}_{\leq a_1+a_1}+\underbrace{a_2+a_4+a_4+a_4}_{\leq a_2+a_2+a_3+a_3}+\cdots +\underbrace{a_{2^n}+a_{2^{n+1}}+\cdots +a_{2^{n+1}}}_{\leq a_{2^n}+a_{2^n}+a_{(2^n+1)}+a_{(2^n+1)}+\cdots +a_{(2^{n+1}-1)}}+\cdots \\
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| & \leq a_1 + a_1 + a_2 +a_2 + a_3 + a_3 + \cdots + a_n + a_n + \cdots = 2 \sum_{n=1}^{\infty} a_n.
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| \end{align}</math>
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| And we have convergence, again by direct comparison. And we are done. Note that we have obtained the estimate
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| :<math>\sum_{n=1}^{\infty} a_n \leq \sum_{n=0}^{\infty} 2^n a_{2^n} \leq 2 \sum_{n=1}^{\infty} a_n.</math>
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| This proof is a generalization of [[Nicole Oresme|Oresme's]] proof of the divergence of the [[harmonic series (mathematics)|harmonic series]].
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| == Generalizations ==
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| The following generalization is due to [[Schlömilch]]. Let <math>\sum_{n=0}^{\infty} a_n</math> be an infinite real series whose terms are positive and non-increasing, and let <math>u_0<u_1<u_2<\cdots</math> be a strictly increasing sequence of positive integers such that
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| :<math>\frac{\Delta u_n}{\Delta u_{n-1}} = \frac{u_{n+1}-u_n}{u_n-u_{n-1}}</math>
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| is bounded, where <math>\Delta u_n</math> is the [[Finite difference|forward difference]]. Then the series <math>\sum_{n=0}^{\infty} a_n</math> converges if the series
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| :<math>\sum_{n=0}^{\infty} {\Delta u_n} a_{u_n} = \sum_{n=0}^{\infty} (u_{n+1}-u_n) a_{u_n}</math>
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| converges.
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| Taking <math>u_n = 2^n</math>, we see <math>\Delta u_n = 2^n</math>, so the Cauchy condensation test emerges as a special case.
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| == References ==
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| * Bonar, Khoury (2006). ''Real Infinite Series''. Mathematical Association of America. ISBN 0-88385-745-6.
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| == External links ==
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| * [http://pirate.shu.edu/projects/reals/numser/t_conden.html Cauchy condensation test proof]
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| * [http://www.encyclopediaofmath.org/index.php/Cauchy_test Cauchy test] at [http://www.encyclopediaofmath.org/ Encyclopedia of Mathematics]
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| [[Category:Mathematical series]]
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| [[Category:Convergence tests]]
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Hello. Let me introduce the author. Her title is Emilia Shroyer but it's not the most female title out there. What I love performing is taking part in baseball but I haven't produced a dime with it. He used to be unemployed but now he is a meter reader. Her spouse and her reside in Puerto Rico but she will have to transfer 1 day or an additional.
Feel free to visit my web page - kard.dk