# Comparing two binomial coefficient sums

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.

• motivation?${}$ – Cheerful 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

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).
• Thank you. Could you please explain how did you get term $(2j-q)^2$? – David Apr 19 '12 at 4:42