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Let $j \in N, n\in N, n>1, q\geq 2$. I would like to show that $$ \sum_{j=\frac{n}{\ln n}}^{\sqrt n/2}(2j-n)^q{n \choose j}<\sum_{j=\sqrt n/2}^{{\frac n2}}(2j-n)^q{n \choose j} $$

Any help would be appreciated.

Thank you.

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motivation?${}$ –  Grumpy Parsnip Apr 19 '12 at 4:16
    
I am working under some approximation question. I wanted to 'cut out' part of the sum. It seems that the LHS sum is much more smaller then the RHS sum. –  David Apr 19 '12 at 4:20
    
Are you sure about the limits of the LHS? –  Yuval Filmus Apr 19 '12 at 4:28
    
cross-posted to MO: mathoverflow.net/questions/94489/one-sum-less-then-another –  user16299 Jun 8 '12 at 21:21

1 Answer 1

up vote 2 down vote accepted

One approach would be to bound the term $(2j-n)^2$ on both sides, use the upper bound on the LHS, and the upper bound on the RHS. Then estimate the binomial sums using the identities $$ \left( \frac{n}{k} \right)^k \leq \binom{n}{k} \leq n^k, $$ or if you want to be fancier, using Stirling's approximation or even using Berry-Esséen (recall that a binomial distribution is close to a normal distribution).

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Thank you. Could you please explain how did you get term $(2j-q)^2$? –  David Apr 19 '12 at 4:42
    
That was just a typo, sorry. –  Yuval Filmus Apr 19 '12 at 15:07

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