**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.

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**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.

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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.

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

u spelled "iff" wrong

ReplyDeleteiff = if and only if :P

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

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

ReplyDeleteAnyway, Cauchy condensation gives

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

Which clearly converges by ratio test.

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

ReplyDelete