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Let $n_i$, $i=1,\ldots,m+1$ be nonnegative natural numbers, sum of which $\sum_{i=1}^{m+1}n_i=N$.

I woul like to find an upper bound for the following$$ \sum_{i=1}^{m+1}\frac{\sqrt n_i}{2^{i-1}}$$

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How about $\sum_{i=1}^{m+1} \frac{\sqrt{N}}{2^{i-1}}$? –  Aryabhata Apr 13 '12 at 5:36
    
@Aryabhata: would it be true? In this case, using geometric progression, we get bound $\frac{\sqrt N}{2^{m+1}}$. Now comparig with result in the answer... –  David Apr 13 '12 at 6:06
    
@David: I think it is: $\sqrt{n_i} \le \sqrt{N}$. –  Aryabhata Apr 13 '12 at 7:14

1 Answer 1

up vote 1 down vote accepted

Using $x=(\sqrt{n_1},...,\sqrt{n_{m+1}})$, $y=(1,\frac{1}{2},...,\frac{1}{2^m})$, the Cauchy-Schwarz inequality ($<x,y> \leq ||x|| ||y||$) gives $$\sum_{i=1}^{m+1}\frac{\sqrt n_i}{2^{i-1}} \leq \sqrt{N} \frac{2}{\sqrt{3}} \sqrt{1-\frac{1}{4^{m+1}}}$$

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