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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

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

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

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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}

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  • $\begingroup$ 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
    Feb 7, 2012 at 16:46
  • $\begingroup$ (it is not, indeed, as remarked in Suvrit's link) $\endgroup$
    – Pietro Majer
    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
    Feb 7, 2012 at 16:59
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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)

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