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Inmathematics, theCauchy condensation test, named afterAugustin-Louis Cauchy, is a standardconvergence test forinfinite series. For anon-increasingsequence${\displaystyle f(n)}$ of non-negativereal numbers, the series$\sum _{n=1}^{\mathrm{\infty }}f(n)$converges if and only if the "condensed" series$\sum _{n=0}^{\mathrm{\infty }}{2}^{n}f(2^n)$ converges. Moreover, if they converge, the sum of the condensed series is no more than twice as large as the sum of the original.

The Cauchy condensation test follows from the stronger estimate,$${\displaystyle \sum _{n=1}^{\mathrm{\infty }}f(n)\le \sum _{n=0}^{\mathrm{\infty }}{2}^{n}f(2^n)\le 2\sum _{n=1}^{\mathrm{\infty }}f(n),}$$which should be understood as aninequality ofextended real numbers. The essential thrust of aproof follows, patterned afterOresme's proof of thedivergence of theharmonic series.

To see the first inequality, the terms of the original series are rebracketed into runs whose lengths arepowers 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.

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$2\sum _{n=1}^{\mathrm{\infty }}f(n)$ whichbegins with$f(2^n)$ lines up with the end of the run of$\sum _{n=0}^{\mathrm{\infty }}{2}^{n}f(2^n)$ whichends with$f(2^n)$, so that the former stays always "ahead" of the latter.

The "condensation" transformation$f(n)\to 2^{n}f(2^n)$ recalls theintegral variable substitution$x\to e^x$ yielding$f(x)\mathrm{d}x\to e^{x}f(e^x)\mathrm{d}x$.

Pursuing this idea, theintegral test for convergence gives us, in the case ofmonotone${\displaystyle f}$, that$\sum _{n=1}^{\mathrm{\infty }}f(n)$ converges if and only if${\displaystyle {\displaystyle {\int }_1^{\mathrm{\infty }}f(x)\mathrm{d}x}}$ converges. The substitution$x\to 2^x$ yields the integral${\displaystyle {\displaystyle \log 2{\int }_2^{\mathrm{\infty }}{2}^{x}f(2^x)\mathrm{d}x}}$. We then notice that, where the right hand side comes from applying the integral test to the condensed series$\sum _{n=0}^{\mathrm{\infty }}{2}^{n}f(2^n)$. Therefore,$\sum _{n=1}^{\mathrm{\infty }}f(n)$ converges if and only if$\sum _{n=0}^{\mathrm{\infty }}{2}^{n}f(2^n)$ converges.

The test can be useful for series wheren appears as in a denominator inf. For the most basic example of this sort, the harmonic series$\sum _{n=1}^{\mathrm{\infty }}1/n$ is transformed into the series$\sum 1$, which clearly diverges.

For a more complex example, take$${\displaystyle f(n):=n^{-a}(\log n{)}^{-b}(\log \log n{)}^{-c}.}$$

Here the series definitely converges fora > 1, and diverges fora < 1. Whena = 1, the condensation transformation gives the series$${\displaystyle \sum n^{-b}(\log n{)}^{-c}.}$$

Thelogarithms "shift to the left". So whena = 1, we have convergence forb > 1, divergence forb < 1. Whenb = 1 the value ofc enters.

This result readily generalizes: the condensation test, applied repeatedly, can be used to show that for${\displaystyle k=1,2,3,\dots }$, the generalized Bertrand series$${\displaystyle \sum _{n\ge N}\frac{1}{n\cdot \log n\cdot \log \log n\cdots {\log }^{\circ (k-1)}n\cdot ({\log }^{\circ k}n{)}^{\alpha }}(N=\lfloor {\exp }^{\circ k}(0)\rfloor +1)}$$ converges for${\displaystyle \alpha >1}$ and diverges for.^[1] Here${\displaystyle f^{\circ m}}$ denotes themthiterate of a function${\displaystyle f}$, so thatThe lower limit of the sum,${\displaystyle N}$, 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${\displaystyle k=2}$ and${\displaystyle \alpha =1}$, the partial sum exceeds 10 only after${\displaystyle {10}^{{10}^{100}}}$(agoogolplex) terms; yet the series diverges nevertheless.

A generalization of the condensation test was given byOskar Schlömilch.^[2] Letu(n) be a strictly increasing sequence of positiveintegers such that the ratio of successivedifferences is bounded: there is a positive real numberN, for which

Then, provided that${\displaystyle f(n)}$ meets the same preconditions as inCauchy's convergence test, the convergence of the series$\sum _{n=1}^{\mathrm{\infty }}f(n)$ is equivalent to the convergence of$${\displaystyle \sum _{n=0}^{\mathrm{\infty }}\mathrm{\Delta }u(n)f(u(n))=\sum _{n=0}^{\mathrm{\infty }}(u(n+1)-u(n))f(u(n)).}$$

Taking$u(n)=2^n$ so that$\mathrm{\Delta }u(n)=u(n+1)-u(n)=2^n$, the Cauchy condensation test emerges as a special case.

• Bonar, Khoury (2006).Real Infinite Series. Mathematical Association of America.ISBN 0-88385-745-6.

• Cauchy condensation test proof

• Augustin-Louis Cauchy
• Convergence tests

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