Mertens theorem (Cauchy product)

The set of Mertens (by Franz Mertens ) is a mathematical theorem from the analysis , a statement about convergence of a Cauchy-product of two rows supplies.

formulation

If and are convergent series, where at least one of the two converges absolutely , then the Cauchy product , where is, converges to . ${\displaystyle \textstyle A=\sum _{k=0}^{\infty }a_{k}}$ ${\displaystyle \textstyle B=\sum _{k=0}^{\infty }b_{k}}$ ${\displaystyle \textstyle \sum _{k=0}^{\infty }c_{k}}$ ${\displaystyle \textstyle c_{k}=\sum _{j=0}^{k}a_{j}b_{k-j}}$ ${\displaystyle AB}$

proof

Without restriction, let the series be absolutely convergent. To show now is that the partial sum to converge. ${\displaystyle A}$ ${\displaystyle \textstyle S_{n}=\sum _{k=0}^{n}c_{k}}$ ${\displaystyle AB}$

In the following be and . ${\displaystyle \textstyle A_{n}:=\sum _{k=0}^{n}a_{k}}$ ${\displaystyle \textstyle B_{n}=\sum _{k=0}^{n}b_{k}}$

Cauchy product

1. ${\displaystyle AB}$ can be written as ${\displaystyle \textstyle (A-A_{n})B+\sum _{k=0}^{n}a_{k}B}$
2. ${\displaystyle S_{n}}$ can be written as ${\displaystyle \textstyle \sum _{k=0}^{n}a_{k}B_{n-k}}$

The difference formation 1. - 2. gives

${\displaystyle AB-S_{n}=(A-A_{n})B+\sum _{k=0}^{n}a_{k}(B-B_{n-k})}$

It converges to zero and the last row can be split with ${\displaystyle (A-A_{n})B}$ ${\displaystyle N:=\left\lfloor {\tfrac {n}{2}}\right\rfloor }$

${\displaystyle \underbrace {\sum _{k=0}^{N}a_{k}(B-B_{n-k})} _{=P_{n}}+\underbrace {\sum _{k=N+1}^{n}a_{k}(B-B_{n-k})} _{=Q_{n}}\,.}$

It applies

${\displaystyle |P_{n}|\leq \sum _{k=0}^{N}|a_{k}|\cdot |B-B_{n-k}|\leq \max _{N\leq k\leq n}|B-B_{k}|\,\sum _{k=0}^{N}|a_{k}|\to 0\,,}$

for the last term is a product of a null sequence with a bounded sequence. Since the null sequence has to be restricted, there is a with . thats why ${\displaystyle (B-B_{k})}$ ${\displaystyle C>0}$ ${\displaystyle |B-B_{k}|<C\quad \forall k\in \mathbb {N} }$

${\displaystyle |Q_{n}|\leq \sum _{k=N+1}^{n}|a_{k}|\cdot |B-B_{n-k}|\leq C\sum _{k=N+1}^{n}|a_{k}|\to 0}$

according to the Cauchy criterion . So what follows immediately applies . ${\displaystyle AB-S_{n}\to 0}$ ${\displaystyle S_{n}\to AB}$

The Cauchy product under conditional convergence

If both output series only converge to a limited extent, then the Cauchy product does not have to converge, as the example shows: The Cauchy product of the series with does not converge, see Cauchy product formula # A divergent series . ${\displaystyle A,B}$ ${\displaystyle a_{k}=b_{k}={\tfrac {(-1)^{k}}{\sqrt {k+1}}}}$

Hardy showed, however, that the Cauchy product also converges for two series which are only partially convergent if the sequences and are limited. For the known not absolutely convergent output series ${\displaystyle a_{k}\cdot k}$ ${\displaystyle b_{k}\cdot k}$

${\displaystyle A=B=\sum _{n=0}^{\infty }(-1)^{n}{\frac {1}{2n+1}}}$

with value the Cauchy product is thus convergent with value . ${\displaystyle {\tfrac {\pi }{4}}}$ ${\displaystyle {\tfrac {\pi ^{2}}{16}}}$

Individual evidence

1. ^ Konrad Königsberger: Analysis 1 - 5th edition. Springer-Verlag Berlin Heidelberg New York, ISBN 3-540-41282-4 ; P. 74 (end of section 6.3)
2. ^ The Multiplication of Conditionally Convergent Series, GH Hardy, Proceedings of the London Mathematical Society, Volume s2-6, Issue 1, 1908, Pages 410-423, https://doi.org/10.1112/plms/s2-6.1.410