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 $x\_k = 1/k\;\;$ ($k = 1, 2, ...$).



What bounds the ratio of summed partial harmonic means to a sum?
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](http://en.wikipedia.org/wiki/Hardy%27s_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](http://en.wikipedia.org/wiki/Carleman%27s_inequality).


