Tuesday, March 27, 2007

Condensation Sensation. Topic: Calculus/S&S.

Theorem: (Cauchy Condensation Test) If $ \displaystyle \{a_n\}_{n = 0}^{\infty} $ is a monotonically decreasing sequence of positive reals and $ p $ is a positive integer, then

$ \displaystyle \sum_{n=0}^{\infty} a_n $ converges if and only if $ \displaystyle \sum_{n=0}^{\infty} p^n a_{p^n} $ converges.


Problem: Determine the convergence of $ \displaystyle \sum_{n=2}^{\infty} \frac{1}{(\ln{n})^k} $, where $ k $ is a positive real.

Solution: Well, let's apply the Cauchy condensation test. Then we know that

$ \displaystyle \sum_{n=2}^{\infty} \frac{1}{(\ln{n})^k} $

converges if and only if

$ \displaystyle \sum_{n=2}^{\infty} \frac{p^n}{(\ln{p^n})^k} = \sum_{n=2}^{\infty} \frac{p^n}{(n \ln{p})^k} = \frac{1}{(\ln{p})^k} \sum_{n=2}^{\infty} \frac{p^n}{n^k} $

does. But this clearly diverges due to the fact that the numerator is exponential and the denominator is a power function. QED.


Comment: This is a pretty powerful test for convergence, at least in the situations in which it can be applied. The non-calculus proof for the divergence of the harmonic series is very similar to the Cauchy condensation test; in fact, the condensation test would state that

$ \displaystyle \sum_{n=1}^{\infty} \frac{1}{n} $ converges iff $ \displaystyle \sum_{n=1}^{\infty} \frac{p^n}{p^n} $ converges,

which clearly shows that the harmonic series diverges.


Practice Problem: Determine the convergence of $ \displaystyle \sum_{n=2}^{\infty} \frac{1}{(\ln{n})^{\ln{n}}} $.


  1. iff = if and only if :P

  2. Hmm. I was going to ask for a proof, but it occurred to me that (condensed series)

  3. Oops. I wanted to write (condensed series) ≤ (original series) ≤ p * (condensed series). Yeah.

    Anyway, Cauchy condensation gives

    $ \displaystyle \sum_{n=2}^{\infty} \frac{p^n}{(n \ln p)^{n \ln p} } $

    Which clearly converges by ratio test.

  4. The root test is also a good way to go to show convergence of that last thing.