4
$\begingroup$

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).

Questions: What is the name of the above inequality? Is $C=e$ the best constant for it? Is it attained?

$$*$$ edit. (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

$$MG(a_1,\dots,a_n) \le \frac{e}{n(n+1)}\, \sum_{1\le k \le n} k a_k \, ,$$

whence

$$\sum_{n\ge1}MG(a_1,\dots,a_n) \le\, e\, \sum_{k\ge1} \bigg( \sum_{n\ge k} \frac{1}{n(n+1)}\bigg) \, k a_k =\, e\, \sum_{k\ge1}\, a_k \, .$$

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ I suddenly remembered the name of this inequality, and have now updated my answer to included the details :-) $\endgroup$
    – Suvrit
    Commented Feb 7, 2012 at 16:40

2 Answers 2

Reset to default
2
$\begingroup$

You can’t have $C< e$. Fix $N$, and define $$a_n=\begin{cases}\tfrac1n&n\le N\\\\\\\\0&\text{otherwise.}\end{cases}$$ 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}

Improve this answer
$\endgroup$
3
  • $\begingroup$ Excellent. Note that the proof of the inequality (that I now added in an edit above) uses the inequality $MG\le MA$ for the $n$-ples $(1a_1,…,na_n)$, which is an equality exactly when $ka_k$ is constant. So now I think this should have suggested me trying the truncated of the harmonic series as you did. It remains the curiosity to know whether or not there is a series realizing the best constant e (I think not). $\endgroup$
    – Pietro Majer
    Commented Feb 7, 2012 at 16:46
  • $\begingroup$ (it is not, indeed, as remarked in Suvrit's link) $\endgroup$
    – Pietro Majer
    Commented Feb 7, 2012 at 16:55
  • $\begingroup$ Indeed, and this is easy to see from your proof: the inequality $(n!)^{-(1/n)}< e/(n+1)$ is strict, hence you get $\sum_nMG(a_1,\dots,a_n)< e\sum_na_n$ as long as at least one of the terms $MA(a_1,\dots,na_n)$ is nonzero, i.e., unless $a_n$ is the constant $0$ sequence. $\endgroup$
    – Emil Jeřábek
    Commented Feb 7, 2012 at 16:59
4
$\begingroup$

Your alleged inequality is the "well-known" Carleman's inequality, for which it is known that $C=e$ is the best constant.

There are several interesting generalizations to this basic inequality; the wikipedia page lists some. Also, one proof of this inequality follows directly from Hardy's inequality.

EDIT. You might also enjoy the survey: Carleman's inequality: history and new generalizations by J. Pečarić (Aequationes Mathematicae, Volume 61, Numbers 1-2, 49-62)

Improve this answer
$\endgroup$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.