Carleman inequality

The Carleman inequality , named after the Swedish mathematician Torsten Carleman , is an elementary inequality in analysis . It says that a series of geometric means of a sequence is bounded by a constant multiple of the series from above. More precisely, it says that Euler's number is the smallest constant which, as a multiple, fulfills this limit. ${\displaystyle (a_{k})_{k}}$ ${\displaystyle \sum a_{k}}$ ${\displaystyle e}$

The Carleman inequality was first published in 1923 by Torsten Carleman.

sentence

statement

Be a sequence of real, non-negative numbers. Denote Euler's number . Then: ${\displaystyle (a_{k})_{k}=(a_{1},a_{2},a_{3},\ldots )}$ ${\displaystyle e}$ ${\displaystyle e\approx 2{,}71828\ldots }$

${\displaystyle \sum _{k=1}^{\infty }{\sqrt[{k}]{(a_{1}a_{2}\ldots a_{k})}}\leq e\cdot \sum _{k=1}^{\infty }a_{k}~\,}$.

Where is the smallest number that fulfills this statement. ${\displaystyle e}$

proof

Because is ( telescope sum ) ${\displaystyle {\frac {1}{n(n+1)}}={\frac {1}{n}}-{\frac {1}{n+1}}}$ ${\displaystyle \sum _{n=k}^{\infty }{\frac {1}{n(n+1)}}={\frac {1}{k}}\quad }$

and it follows ${\displaystyle {\frac {1}{e^{n}}}<\prod _{k=1}^{n}\left({\frac {k}{k+1}}\right)^{k}=\prod _{k=1}^{n}{\frac {(k+1)k^{k}}{(k+1)^{k+1}}}={\frac {(n+1)!}{(n+1)^{n+1}}}={\frac {n!}{(n+1)^{n}}}}$ ${\displaystyle {\frac {1}{e}}<{\frac {\sqrt[{n}]{n!}}{n+1}}}$

${\displaystyle \sum _{k=1}^{\infty }a_{k}=\sum _{k=1}^{\infty }\sum _{n=k}^{\infty }{\frac {1}{n(n+1)}}k\,a_{k}=\sum _{1\leq k\leq n}{\frac {1}{n(n+1)}}k\,a_{k}=\sum _{n=1}^{\infty }{\frac {1}{n+1}}\;{\frac {1}{n}}\sum _{k=1}^{n}k\,a_{k}}$and that is according to the AM-GM inequality

${\displaystyle \geq \sum _{n=1}^{\infty }{\frac {1}{n+1}}{\sqrt[{n}]{\prod _{k=1}^{n}(k\,a_{k})}}=\sum _{n=1}^{\infty }{\frac {\sqrt[{n}]{n!}}{n+1}}{\sqrt[{n}]{\prod _{k=1}^{n}a_{k}}}\geq {\frac {1}{e}}\sum _{n=1}^{\infty }{\sqrt[{n}]{a_{1}\cdots a_{n}}}\qquad \Box }$

variants

The following continuous variant of the Carleman inequality applies to a function with : ${\displaystyle f}$ ${\displaystyle f\not \equiv 0}$

${\displaystyle \int _{0}^{\infty }\exp \left({\frac {1}{x}}\int _{0}^{x}\ln f(t)\,dt\right)\,dx<e\cdot \int _{0}^{\infty }f(x)\,dx\,}$.

literature

• GH Hardy , JE Littlewood , G. Pólya : Inequalities. 2nd edition. Cambridge University Press, Cambridge et al. 1952 (2nd edition, 1st paperback edition, reprinted. Transferred to digital print. Ibid 2001, ISBN 0-521-35880-9 ).