Английская Википедия:Cauchy condensation test
Шаблон:Short description Шаблон:Distinguish Шаблон:Calculus
In mathematics, the Cauchy condensation test, named after Augustin-Louis Cauchy, is a standard convergence test for infinite series. For a non-increasing sequence <math>f(n)</math> of non-negative real numbers, the series <math display="inline">\sum\limits_{n=1}^{\infty} f(n)</math> converges if and only if the "condensed" series <math display="inline">\sum\limits_{n=0}^{\infty} 2^{n} f(2^{n})</math> converges. Moreover, if they converge, the sum of the condensed series is no more than twice as large as the sum of the original.
Estimate
The Cauchy condensation test follows from the stronger estimate, <math display="block"> \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> which should be understood as an inequality of extended real numbers. The essential thrust of a proof follows, patterned after Oresme's proof of the divergence of the harmonic series.
To see the first inequality, the terms of the original series are rebracketed into runs whose lengths are powers of two, and then each run is bounded above by replacing each term by the largest term in that run. That term is always the first one, since by assumption the terms are non-increasing. <math display="block">\begin{array}{rcccccccl}\displaystyle \sum\limits_{n=1}^{\infty} f(n) & = &f(1) & + & f(2) + f(3) & + & f(4) + f(5) + f(6) + f(7) & + & \cdots \\
& = &f(1) & + & \Big(f(2) + f(3)\Big) & + & \Big(f(4) + f(5) + f(6) + f(7)\Big) & + &\cdots \\
& \leq &f(1) & + & \Big(f(2) + f(2)\Big) & + & \Big(f(4) + f(4) + f(4) + f(4)\Big) & + &\cdots \\
& = &f(1) & + & 2 f(2) & + & 4 f(4)& + &\cdots = \sum\limits_{n=0}^{\infty} 2^{n} f(2^{n})
\end{array}</math>
To see the second inequality, these two series are again rebracketed into runs of power of two length, but "offset" as shown below, so that the run of <math display="inline"> 2 \sum_{n=1}^{\infty} f(n)</math> which begins with <math display="inline"> f(2^{n})</math> lines up with the end of the run of <math display="inline"> \sum_{n=0}^{\infty} 2^{n} f(2^{n})</math> which ends with <math display="inline"> f(2^{n})</math>, so that the former stays always "ahead" of the latter. <math display="block">\begin{align} \sum_{n=0}^{\infty} 2^{n}f(2^{n}) & = f(1) + \Big(f(2) + f(2)\Big) + \Big(f(4) + f(4) + f(4) +f(4)\Big) + \cdots \\ & = \Big(f(1) + f(2)\Big) + \Big(f(2) + f(4) + f(4) + f(4)\Big) + \cdots \\ & \leq \Big(f(1) + f(1)\Big) + \Big(f(2) + f(2) + f(3) + f(3)\Big) + \cdots = 2 \sum_{n=1}^{\infty} f(n) \end{align}</math>
Integral comparison
The "condensation" transformation <math display="inline"> f(n) \rarr 2^{n} f(2^{n})</math> recalls the integral variable substitution <math display="inline"> x \rarr e^{x}</math> yielding <math display="inline"> f(x)\,\mathrm{d}x \rarr e^{x} f(e^{x})\,\mathrm{d}x</math>.
Pursuing this idea, the integral test for convergence gives us, in the case of monotone <math>f</math>, that <math display="inline">\sum\limits_{n=1}^{\infty}f(n)</math> converges if and only if <math>\displaystyle\int_{1}^{\infty}f(x)\,\mathrm{d}x</math> converges. The substitution <math display="inline"> x\rarr 2^x</math> yields the integral <math>\displaystyle \log 2\ \int_{2}^{\infty}\!2^{x}f(2^{x})\,\mathrm{d}x</math>. We then notice that <math>\displaystyle \log 2\ \int_{2}^{\infty}\!2^{x}f(2^{x})\,\mathrm{d}x < \log 2\ \int_{0}^{\infty}\!2^{x}f(2^{x})\,\mathrm{d}x</math>, where the right hand side comes from applying the integral test to the condensed series <math display="inline">\sum\limits_{n=0}^{\infty} 2^{n}f(2^{n})</math>. Therefore, <math display="inline">\sum\limits_{n=1}^{\infty} f(n)</math> converges if and only if <math display="inline">\sum\limits_{n=0}^{\infty} 2^{n}f(2^{n})</math> converges.
Examples
The test can be useful for series where Шаблон:Mvar appears as in a denominator in Шаблон:Math. For the most basic example of this sort, the harmonic series <math display="inline">\sum_{n=1}^{\infty} 1/n</math> is transformed into the series <math display="inline">\sum 1</math>, which clearly diverges.
As a more complex example, take <math display="block">f(n) := n^{-a} (\log n)^{-b} (\log \log n)^{-c}.</math>
Here the series definitely converges for Шаблон:Math, and diverges for Шаблон:Math. When Шаблон:Math, the condensation transformation gives the series <math display="block">\sum n^{-b} (\log n)^{-c}.</math>
The logarithms "shift to the left". So when Шаблон:Math, we have convergence for Шаблон:Math, divergence for Шаблон:Math. When Шаблон:Math the value of Шаблон:Mvar enters.
This result readily generalizes: the condensation test, applied repeatedly, can be used to show that for <math>k = 1,2,3,\ldots</math>, the generalized Bertrand series <math display="block">\sum_{n\geq N} \frac{1}{n \cdot \log n \cdot \log\log n \cdots \log^{\circ (k-1)} n \cdot(\log^{\circ k} n)^\alpha} \quad\quad (N=\lfloor \exp^{\circ k} (0) \rfloor+1)</math> converges for <math>\alpha > 1</math> and diverges for <math>0 < \alpha \leq 1</math>.[1] Here <math>f^{\circ m}</math> denotes the Шаблон:Mathth iterate of a function <math>f</math>, so that <math display="block"> f^{\circ m} (x) := \begin{cases} f(f^{\circ(m-1)}(x)), & m=1, 2, 3,\ldots; \\ x, & m = 0. \end{cases}</math> The lower limit of the sum, <math>N</math>, was chosen so that all terms of the series are positive. Notably, these series provide examples of infinite sums that converge or diverge arbitrarily slowly. For instance, in the case of <math>k = 2</math> and <math>\alpha = 1</math>, the partial sum exceeds 10 only after <math>10^{10^{100}}</math>(a googolplex) terms; yet the series diverges nevertheless.
Schlömilch's generalization
A generalization of the condensation test was given by Oskar Schlömilch.[2] Let Шаблон:Math be a strictly increasing sequence of positive integers such that the ratio of successive differences is bounded: there is a positive real number Шаблон:Math, for which <math display="block">{\Delta u(n) \over \Delta u(n{-}1)} \ =\ {u(n{+}1)-u(n) \over u(n)-u(n{-}1)} \ <\ N \ \text{ for all } n.</math>
Then, provided that <math>f(n)</math> meets the same preconditions as in Cauchy's convergence test, the convergence of the series <math display="inline">\sum_{n=1}^{\infty} f(n)</math> is equivalent to the convergence of <math display="block">\sum_{n=0}^{\infty} {\Delta u(n)}\, f(u(n)) \ =\ \sum_{n=0}^{\infty} \Big(u(n{+}1)-u(n)\Big) f(u(n)).</math>
Taking <math display="inline">u(n) = 2^n</math> so that <math display="inline"> \Delta u(n) = u(n{+}1)-u(n) = 2^n</math>, the Cauchy condensation test emerges as a special case.
References
- Bonar, Khoury (2006). Real Infinite Series. Mathematical Association of America. Шаблон:ISBN.
External links
- ↑ Шаблон:Cite book
- ↑ Elijah Liflyand, Sergey Tikhonov, & Maria Zeltse (2012) Extending tests for convergence of number series page 7/28 via Brandeis University
