Cauchy's compression criterion

The Cauchy criterion compression, also known as Cauchy's compression rate, compression principle, dilution rate or condensation criterion (after Augustin-Louis Cauchy ) is a mathematical convergence criterion , ie a means of deciding whether a infinite series convergent or divergent.

formulation

Let be a monotonically decreasing sequence of non-negative real numbers. Then has the infinite series ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$

${\displaystyle S=\sum _{n=1}^{\infty }a_{n}}$

the same convergence behavior as the condensed series

${\displaystyle T=\sum _{k=0}^{\infty }2^{k}a_{2^{k}}}$,

that is, one series converges if and only if the other converges.

Evidence sketch

The effect of this criterion can be thought of as a consideration of the upper and lower sums of the series to be examined. The sequence is divided into blocks of increasing length and estimated against maximum and minimum in each block. Since the sequence was assumed to be monotonically falling, the maximum is identical to the first and the minimum to the last term of each block. ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$

The criterion now results from the majorant criterion . The most common block division is that according to powers of two with blocks . To prove convergence, the majorante is constructed ${\displaystyle a_{2^{k}},a_{2^{k}+1},\dots ,a_{2^{k+1}-1}}$ ${\displaystyle (b_{n})_{n\in \mathbb {N} }}$

${\displaystyle b_{2^{k}+m}:=a_{2^{k}}\geq a_{2^{k}+m}}$for . ${\displaystyle 0\leq m<2^{k}}$

For every index k the majorante contains terms with the same value , so the majorante converges exactly when it converges. ${\displaystyle 2^{k}}$ ${\displaystyle a_{2^{k}}}$ ${\displaystyle \textstyle T=\sum _{k=0}^{\infty }2^{k}a_{2^{k}}}$

In order to prove divergence, the minorant is constructed through ${\displaystyle (b_{n})_{n\in \mathbb {N} }}$

${\displaystyle b_{2^{k}+m}:=a_{2^{k+1}}\leq a_{2^{k+1}-1}\leq a_{2^{k}+m}}$for . ${\displaystyle 0\leq m<2^{k}}$

For every index k , the minorant contains terms with the same value , so the minorant diverges precisely when it diverges. ${\displaystyle 2^{k}}$ ${\displaystyle a_{2^{k+1}}}$ ${\displaystyle \textstyle {\frac {1}{2}}(T-a_{1})=\sum _{k=0}^{\infty }2^{k}a_{2^{k+1}}}$

Application example

One application is in the general harmonic series . For a solid hat ${\displaystyle \alpha >0}$

${\displaystyle S=\sum _{n=1}^{\infty }{\frac {1}{n^{\alpha }}}}$

the same convergence behavior as

${\displaystyle T=\sum _{k=0}^{\infty }2^{k}{\frac {1}{(2^{k})^{\alpha }}}=\sum _{k=0}^{\infty }(2^{1-\alpha })^{k}}$.

${\displaystyle T}$is obviously a geometric series with a factor . From their convergence behavior it follows that there is convergence, otherwise divergence. Note the change in the starting value and the index of the series from to . ${\displaystyle q=2^{1-\alpha }}$ ${\displaystyle \alpha >1}$ ${\displaystyle n=1}$ ${\displaystyle k=0}$

The result is analogous for the even slower converging or diverging series

${\displaystyle \sum _{n=2}^{\infty }{\frac {1}{n\cdot \ln(n)^{\alpha }}},\quad \sum _{n=\lceil e\rceil }^{\infty }{\frac {1}{n\cdot \ln(n)\cdot \ln(\ln(n))^{\alpha }}},\quad \sum _{n=\lceil e^{e}\rceil }^{\infty }{\frac {1}{n\cdot \ln(n)\cdot \ln(\ln(n))\cdot \ln(\ln(\ln(n)))^{\alpha }}}\;\cdots }$

for convergence, otherwise divergence. ${\displaystyle \alpha >1}$

generalization

Instead of the partial sequence , more general partial sequences can also be used for compression. Let be a monotonically decreasing sequence of nonnegative real numbers. Then has the infinite series ${\displaystyle a_{2^{k}}}$ ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$

${\displaystyle S=\sum _{n=1}^{\infty }a_{n}}$

the same convergence behavior as the condensed series

${\displaystyle T=\sum _{k=0}^{\infty }f(k)a_{f(k)}}$,

where is a strictly monotonically increasing function on the natural numbers, the ${\displaystyle f(k)}$

${\displaystyle 1<\inf \left\lbrace {f(n+1) \over f(n)}:{n\in \mathbb {N} }\right\rbrace \leq \sup \left\lbrace {f(n+1) \over f(n)}:{n\in \mathbb {N} }\right\rbrace <\infty }$

Fulfills.

swell

• Konrad Königsberger : Analysis 1 . Springer-Verlage, Berlin et al., 2004, ISBN 3-540-41282-4 , page 78.