Is there tight estimates for the following logarithmic summation ($\gamma\in(0,1)$)


$$\ln\Bigg(\sum\_{t=\frac{n^{}}2-\gamma n^\gamma}^{\frac{n^{}}2+\gamma n^\gamma}\sum\_{\ell=\frac{n^{}}2-\gamma n^\gamma}^{\frac{n^{}}2+\gamma n^\gamma}\sum\_{k=0}^t\binom{\ell}{k}\binom{n-\ell}{t-k}\Bigg)?$$


Is it roughly bounded above by $$\ln\Bigg(poly(n)\sum\_{k=0}^t\binom{\ell}{k}\binom{n-\ell}{t-k}\Bigg)$$ for some **minimal** choice of $$\sum\_{k=0}^t\binom{\ell}{k}\binom{n-\ell}{t-k}?$$ The summand though seems amenable to Cauchy-Schwarz gives very loose bounds.



Tight estimates for binomial summation
By [Vandermonde's identity](https://en.wikipedia.org/wiki/Vandermonde%27s_identity), $$\sum\_{k=0}^t\binom{\ell}{k}\binom{n-\ell}{t-k}=\binom{n}{t},$$
so the triple sum reduces to
$$
\sum\_{t=n/2-\gamma n^\gamma}^{n/2+\gamma n^\gamma}\sum\_{\ell=n/2-\gamma n^\gamma}^{n/2+\gamma n^\gamma}\binom{n}{t}
=\sum\_{t=n/2-\gamma n^\gamma}^{n/2+\gamma n^\gamma}(2\gamma n^\gamma+1)\binom{n}{t}
=(2\gamma n^\gamma+1)\sum\_{t=n/2-\gamma n^\gamma}^{n/2+\gamma n^\gamma}\binom{n}{t}.
$$
Now $\sum\_t \binom{n}{t} \le 2^n$, so
$$\ln\left(\sum\_{t,\ell,k}\right) =\ln(2\gamma n^\gamma+1)+\ln\left(\sum\_{t=n/2-\gamma n^\gamma}^{n/2+\gamma n^\gamma}\binom{n}{t}\right) \le \ln(2\gamma n^\gamma+1)+n\ln 2.$$


