## 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.

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

1. u spelled "iff" wrong

2. iff = if and only if :P

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

4. 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.

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