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I just finished reading the Wikipedia article on the Cauchy condensation test. I understand the trapezoidal view, but apparently "the 'condensation' of terms is analogous to a substitution of an exponential function". Can anyone explain what is meant by this?

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2 Answers

up vote 5 down vote accepted

Explanation is quite straightforward. Consider integral $\int f(x)dx$. After change of variable $x=2^t$ it becomes $const\cdot\int 2^t f(2^t)dt$ — compare with $\sum 2^n f(2^n)$ from Cauchy condensation test.

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"Condensation" because the series $a_n$ is compressed to a subseries $a_{2^n}$ carrying all the same information about convergence.

The proof of the condensation test is easy, as Grigory M's answer indicates and whenever it works so does a comparison of sum with integral, which is a more fundamental idea. So you might naturally wonder why this seemingly trivial convergence test deserves its own name, other than the fame of its supposed inventor. The reason it is retained in textbooks is that it is often a convenient criterion to use for series when the terms or the partial sums have some dependence on $\log (n)$. The proof of divergence of the harmonic series is an example.

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..or proof of divergence of $\sum\frac{1}{n\log n}$ –  Grigory M Aug 7 '10 at 9:44
Conversely, most of the standard applications of the Integral Test can be done by Condensation. This is convenient if you wish to present series before (improper) integration. In first year calculus this is not often done, but in undergrad analysis this is relatively common: see e.g. Rudin's Principles or math.uga.edu/~pete/243series3.pdf. –  Pete L. Clark Aug 7 '10 at 10:29
By the way, guess who developed the Integral Test (along with, and independently from, Colin Maclaurin)? –  Pete L. Clark Aug 7 '10 at 10:40
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