Bounding the series of the geometric means of the terms of a given positive series - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T20:33:07Z http://mathoverflow.net/feeds/question/87799 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/87799/bounding-the-series-of-the-geometric-means-of-the-terms-of-a-given-positive-serie Bounding the series of the geometric means of the terms of a given positive series Pietro Majer 2012-02-07T15:01:03Z 2012-02-22T04:34:08Z <p>Let $\{ a _ k \} _{k\in\mathbb{N} _ +}$ be a sequence of non-negative numbers, and let $MG(a_1,\dots,a_n)$ denote the geometric mean of the first $n$ terms. Then, the inequality $$\sum _ {n\ge 1}MG(a_1,\dots,a_n) \le C\ \sum _ {n\ge 1} a _ n$$ holds, with $C=e$. This is quite elementary, although not obviously true (for instance, no analogous inequality could hold for the arithmetic means $MA(a_1,a_2,\dots,a_n)$, as the series on the LHS may then diverge even for a converging series on the RHS).</p> <blockquote> <p><strong>Questions:</strong> What is the name of the above inequality? Is $C=e$ the best constant for it? Is it attained?</p> </blockquote> <p>$$*$$ <strong>edit.</strong> (Details on the above inequality). From the Arithmetic-Geometric means inequality $$MG(a_1,\dots,a_n)=MG(1a_1,2a_2,\dots,na_n)(n!)^{-1/n}\le MA(1a_1,2a_2,\dots,na_n)(n!)^{-1/n}\ .$$ Stirling formula in form of inequality, $n!\ge \sqrt{2\pi n}\ n^n e^{-n}$, written for $n+1$, implies $$(n!)^{-1/n} \le \frac{e}{n+1}$$ for all $n\ge1$. So</p> <p>$$MG(a_1,\dots,a_n) \le \frac{e}{n(n+1)}\ \sum_{1\le k \le n} k a_k \ ,$$</p> <p>whence</p> <p>$$\sum_{n\ge1}MG(a_1,\dots,a_n) \le\ e\ \sum_{k\ge1} \bigg( \sum_{n\ge k} \frac{1}{n(n+1)}\bigg) \ k a_k =\ e\ \sum_{k\ge1}\ a_k \ .$$</p> http://mathoverflow.net/questions/87799/bounding-the-series-of-the-geometric-means-of-the-terms-of-a-given-positive-serie/87802#87802 Answer by S. Sra for Bounding the series of the geometric means of the terms of a given positive series S. Sra 2012-02-07T15:19:15Z 2012-02-22T04:34:08Z <p>Your alleged inequality is the "well-known" <a href="http://en.wikipedia.org/wiki/Carleman%27s_inequality" rel="nofollow">Carleman's inequality</a>, for which it is known that $C=e$ is the best constant.</p> <p>There are several interesting generalizations to this basic inequality; the wikipedia page lists some. Also, one proof of this inequality follows directly from <a href="http://en.wikipedia.org/wiki/Hardy%27s_inequality" rel="nofollow">Hardy's inequality</a>.</p> <hr> <p><strong>EDIT.</strong> You might also enjoy the survey: <a href="http://www.springerlink.com/content/ky94cw3r6xl2clfa/" rel="nofollow">Carleman's inequality: history and new generalizations</a> by J. Pečarić (<em>Aequationes Mathematicae, Volume 61, Numbers 1-2, 49-62</em>)</p> http://mathoverflow.net/questions/87799/bounding-the-series-of-the-geometric-means-of-the-terms-of-a-given-positive-serie/87807#87807 Answer by Emil Jeřábek for Bounding the series of the geometric means of the terms of a given positive series Emil Jeřábek 2012-02-07T15:57:22Z 2012-02-07T15:57:22Z <p>You can’t have $C&lt; e$. Fix $N$, and define $$a_n=\begin{cases}\tfrac1n&amp;n\le N\\\\0&amp;\text{otherwise.}\end{cases}$$ Then $$\sum_na_n=H_N=\log N+O(1),$$ and \begin{multline}\sum_n\mathrm{MG}(a_1,\dots,a_n)=\sum_{n=1}^N\frac1{\sqrt[n]{n!}}=\sum_{n=1}^N\frac en\left(1+O\left(\frac{\log n}n\right)\right)\\=eH_N+O(1)=\left(e+O\left(\frac1{\log N}\right)\right)H_N.\end{multline}</p>