I need to find an upper bound for
$\binom{n+\epsilon}{k} \binom{2n-k}{n}$ where $\epsilon>0$ and $k,n$ are positive integers with $0 \leq k \leq n$.
I think an upper bound should be with $k=n/2$ or something around there, not sure how to prove it.
I need to find an upper bound for
$\binom{n+\epsilon}{k} \binom{2n-k}{n}$ where $\epsilon>0$ and $k,n$ are positive integers with $0 \leq k \leq n$.
I think an upper bound should be with $k=n/2$ or something around there, not sure how to prove it.
$a_k:=\binom{n+\epsilon}{k}*\binom{2n-k}{n} =\frac{(n+\epsilon)!}{k!(n+\epsilon-k)!}\frac{(2n-k)!}{n!(n-k)!} $
$\frac{a_{k+1}}{a_k}=\frac{(n+1+\epsilon)!}{k!(n+1+\epsilon-k)!}\frac{(2n+2-k)!}{(n+1)!(n+1-k)!}\frac{k!(n+\epsilon-k)!}{(n+\epsilon)!}\frac{n!(n-k)!}{(2n-k)!} $
$\frac{a_{k+1}}{a_k}=\frac{(n+1+\epsilon)}{(n+1+\epsilon-k)}\frac{(2n+2-k)(2n+1-k)}{(n+1)(n+1-k)}$
Now, solve
$\frac{a_{k+1}}{a_k} \geq 1$
This is quadratic in $k$ and gives you the monotony of $a_k$, which yields the max/min.