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For positive real $x_1$ , $x_2$ ,..., define their $k$th partial harmonic mean as $h_k = k/(1/x_1 +\cdots+1/x_k)$ for $k = 1, 2, ...,$ and let

$\alpha=\sup_{x_1,x_2,... \geqslant0}\: \lim_{n\rightarrow\infty}\dfrac{h_1+\cdots+h_n}{x_1+\cdots+x_n}.$

What is this bound, and for which $x_1$ , $x_2$ ,... is it attained? All I can do is show $\alpha \geqslant 2$ by taking $x_k = 1/k\;\;$ ($k = 1, 2, ...$).

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up vote 5 down vote accepted

This one is an old classic. I think it is due to Hardy, but you should check Hardy-Littlewood-Polya's "Inequalities". This is equivalent to Hardy's inequality with $p=-1$. I think Hardy claimed the result for all $p>1$, and the observation that it holds for all $p\le 0$ as well came a bit later.

The answer is $\alpha=2$ as you suggest. You want to prove (after renaming the variables, sorry!) that $$2(\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n})\geq \sum_{k=1}^n \frac{k}{a_1+\cdots+a_k}$$ and to prove this, the hint is to use induction and prove instead the stronger statement $$2(\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n})\geq \sum_{k=1}^n \frac{k}{a_1+\cdots+a_k}+\frac{n^2}{2(a_1+\cdots+a_n)}$$ now it should be clear. If you consider geometric means instead of harmonic means you get a supremum $\alpha=e$, which is Carleman's inequality.

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Thank you!........ –  John Bentin Jul 9 '11 at 19:32

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