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Let $n_1, \ldots , n_{m+1}$ be natural numbers, possibly zeros. I want to introduce $$ \sum_{i=1}^m \sqrt{\frac{n_i}{n_{i+1}-n_i}} $$ in terms of the s $S=\sum_{i=1}^{m+1}n_i$

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  • $\begingroup$ You mean you want to express the top sum, of square roots of ratios, in terms of $S$? This is not possible if any denominator is zero, of course. $\endgroup$ – bgins Apr 13 '12 at 13:32
  • $\begingroup$ Why do you think there would be such a formula? $\endgroup$ – Thomas Andrews Apr 13 '12 at 13:46
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Both $(n_1,n_2)=(1,4)$ and $(n_1,n_2)=(2,3)$ yield $S=5$ but $\sqrt{\dfrac{n_1}{n_2-n_1}}$ is $\dfrac1{\sqrt3}$ in one case and $\sqrt2$ in the other case. Hence $S$ does not determine the sum you are interested in, even the sequence $(n_i)$ is suitable (that is, either nonnegative and increasing, or nonpositive and decreasing).

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  • $\begingroup$ Thank you, what about $$\sum_{i=1}^{m}\sqrt{\frac{n_i}{n_{i+1}}}$$ it means that this sjm is also impossible to bound in terms of S? $\endgroup$ – Dora Apr 13 '12 at 15:14

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