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.
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.
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).