Math for Non-Geeks/ Cauchy condensation test

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In this chapter, we present the Cauchy condensation test (named by Augustin Louis Cauchy). It allows us to only check the condensed series $\sum _{k=0}^{\mathrm{\infty }}{2}^k{a}_{2^k}$ for convergence , which contains way less elements than the original series $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ . More precisely, in case the series elements $a_k$ are non-negative and decreasing, we know that the original series $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ converges if and only if the condensed series $\sum _{k=0}^{\mathrm{\infty }}{2}^k{a}_{2^k}$ converges. The derivation will involve a direct comparison to a divergent harmonic series $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k}$ and convergence to a convergent generalized harmonic series $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^{\alpha }}$ für $\alpha >1$.

Repetition and derivation of the criterion

For proving divergence of the harmonic series $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k}$ , we used a lower bound for each $2^n$-th partial sum $\sum _{k=1}^{2^n}\frac{1}{k}$ : Template:Math

How can we generalize this concept to a general series $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ ? In order to make the same estimation steps, the series must have some properties identical to the harmonic series:

• The series elements $a_k$ have to be non-negative.
• The sequence of elements $a_k$ has to be monotonically decreasing.

If $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ is a series with non-negative elements fulfilling $a_k\ge a_{k+1}$ for all $k\in \mathbb{N}$. Then Template:Math

So we estimated $\sum _{k=1}^{2^n}{a}_k$ from below by $\frac{1}{2}\sum _{k=0}^n{2}^k{a}_{2^k}$ . That means, that if the condensed series $\sum _{k=0}^{\mathrm{\infty }}{2}^k{a}_{2^k}$ diverges and hence, $\frac{1}{2}\sum _{k=0}^{\mathrm{\infty }}{2}^k{a}_{2^k}$ as well, then the series $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ also diverges by direct comparison. Conversely (by contraposition), if $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ converges, then $\sum _{k=0}^{\mathrm{\infty }}{2}^k{a}_{2^k}$ converges, as well.

For the proof that the generalized harmonic series $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^{\alpha }}$ with $\alpha >1$ converges, we compared the $n$-th partial sum $\sum _{k=1}^n\frac{1}{k^{\alpha }}$ for $n\le 2^{m+1}-1$ to a convergent geometric series $\sum _{i=0}^m{\left(\frac{1}{2^{\alpha -1}}\right)}^i$ . The bounding worked as follows:

We try to do the same for a general series $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ with

• non-negative elements $a_k$
• and $a_{k+1}\le a_k$ for all $k\in \mathbb{N}$ (monotonically decreasing sequence of elements)

Let $n\le 2^{m+1}-1$. then,

So we can also bound $\sum _{k=1}^n{a}_k$ from above by $\sum _{k=0}^n{2}^k{a}_{2^k}$ . Direct comparison can again be applied and leads us to the conclusion: If the condensed series $\sum _{k=0}^{\mathrm{\infty }}{2}^k{a}_{2^k}$ converges, the the original series $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ also converges.

So if elements are non-negative and monotonically decreasing, we have an equivalence between the convergence of the series $\sum _{k=1}^n{a}_k$ and the condensed series $\sum _{k=0}^{\mathrm{\infty }}{2}^k{a}_{2^k}$ . This result is called Cauchy condensation test. It can be very useful, to remove logarithms out of a series. For instance, if $a_k=\ln (ln(k))...$. Then, for the condensed series $a_{2^k}=\ln (k\cdot \ln (2))...$. So condensation can remove double logarithms.

Now, let us formulate these findings in a mathematical language, i.e. a theorem with a proof:

Applications

File:Cauchy Verdichtungssatz für Reihenkonvergenz.webm

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