I have the following expression defined for all $n\geq 1$ and $k\geq 1$:
$$a_{n,k}=\frac{1}{n+1} \binom{2n}{n} - \binom{2n}{n} + \binom{2n}{n-1} + \binom{2n}{n-k-1} - 2\binom{2n}{n-k-2} + \binom{2n}{n-k-3},$$
and I want to compute its asymptotic behavior (actually, the exponential growth would be enough) when $k=n^\alpha$ with $\alpha<1$, say $\alpha=0.99$.
I tried to plug in naively the asymptotic for binomial coefficients but I do not get anything meaningful. Any idea or help?
Using what @Mike Earnest wrote in comment, using the gamma function and simplifying $$a_{n,k}= \binom{2n}{n-k-1} - 2\binom{2n}{n-k-2} + \binom{2n}{n-k-3}$$ $$a_{n,k}=2\, \Gamma (2 n+1)\,\frac{2 k^2+8 k+(7-n) } {\Gamma (n-k)\, \Gamma (n+k+4) }$$
Making $k=n^\alpha$ and expanding as Taylor series around $\alpha=1^-$, you could use
$$\Gamma \left(n-n^{\alpha }\right)=\frac{1}{(1-\alpha ) n \log (n)}+\left(\frac{1}{2 n}-\gamma \right)+O\left((\alpha -1)\right)$$ $$\Gamma \left(n+n^{\alpha }+4\right)=\Gamma (2 n+4)-(1-\alpha) n \log (n) \Gamma (2 n+4) \psi(2 n+4)+O\left((\alpha -1)^2\right)$$ which makes $$\frac 1{\Gamma (n-k)\, \Gamma (n+k+4) }=\frac{ n \log (n)}{\Gamma (2 n+4)}(1-\alpha )+O\left((\alpha -1)\right)$$ $$2 k^2+8 k+(7-n)=\left(2 n^2+7 n+7\right)-4 n (n+2) \log (n)(1-\alpha )+O\left((\alpha -1)^2\right)$$ All of that gives $$a_{n,k}=\frac{ n (2 n^2+7 n+7) \log (n)}{(n+1) (2 n+1) (2 n+3)}(1-\alpha )+O\left((\alpha -1)^2\right)$$ .... which does not look fantastic.