I have to prove that $\sum_{n=l+1}^{\infty} \sqrt{\frac{n}{2^n}} \leq C \sqrt{h\log(\frac{1}{h})}$ where $2^{-l} <h \leq 2^{-l+1}$?.
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Two. Just write it as $(1/2+1/4+\ldots) + (1/4+1/8+\ldots)+(\ldots)+\ldots$ – Peter Franek Nov 21 '15 at 17:57
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@PeterFranek: I want to know the tail sum from $k$. – Sosha Nov 21 '15 at 18:02
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4Same method works as in http://math.stackexchange.com/questions/1330493/how-do-you-prove-sum-frac-n2n-2 – Nov 21 '15 at 18:02
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Marked as duplicate – Varun Iyer Nov 21 '15 at 18:08
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If you start with $k$, just start with the $k$'th of my parenthesis.. what is the sum? – Peter Franek Nov 21 '15 at 21:27
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@PeterFranek: Sorry. I completely missed the square root. P. check – Sosha Nov 22 '15 at 13:46
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1@Sosha With the square root, I doubt that you can find an expression in closed form. (But I don't know) – Peter Franek Nov 22 '15 at 14:33
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@PeterFranek: i posted the original problem. i need upper bound. if you upper bound $\sqrt{n}$ by $n$ then the earlier method work , but it won't give the desired bound, it will give only $\sqrt{h}$ – Sosha Nov 22 '15 at 15:24
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@Sosha I recommend to post a new problem and ask a clear question without changing it then too much. It just needs time to prepare it.. Also, don't forget to include your attempts. – Peter Franek Nov 22 '15 at 18:18