Asymptotic behavior of $\sum_{k = 0}^n q^k \frac{\sqrt{(2n)!}}{(2k)!(n -k)!}$ I would like to understand the asymptotic behavior of $$\sum_{k = 0}^n q^k \frac{\sqrt{(2n)!}}{(2k)!(n -k)!}$$ when $n \to \infty$, depending of $q > 0$.
I did numerical simulations and apparently $$\sum_{k = 0}^n 3^{n -k} \frac{\sqrt{(2n)!}}{(2k)!(n -k)!} \to 0$$ whereas
$$\sum_{k = 0}^n (3/2)^{n -k} \frac{\sqrt{(2n)!}}{(2k)!(n -k)!} \to \infty$$
I tried to used a kind of Stirling formula: $$\frac{n^n}{e^{n -1}} \leq n! \leq \frac{n^{n+1}}{e^{n-1}}$$ but with no sucess.
Do you have any hint?
I am interested in something like : for $q$ big enough, there is a "simple" expression $\epsilon_n \to 0$ such that:
$$\sum_{k = 0}^n q^{n -k} \frac{\sqrt{(2n)!}}{(2k)!(n -k)!} \leq \epsilon_n$$
For the context, I am tryig to understand the "size" of the expansion of $A^k f$ where $f$ is a smooth function and $A(g)(x) = x g(x) - g'(x)$. By using a non commutative binomial expansion, this kind of sum appears.
 A: Denote
$$S(q,n)=\sum_{k = 0}^n q^k \frac{\sqrt{(2n)!}}{(2k)!(n -k)!}.$$
This sum can be expressed in terms of the Kummer $U$-function as follows:
$$
S(q,n)  = ( - 1)^n \frac{{4^n }}{{\sqrt {(2n)!} }}U\!\left( { - n,\frac{1}{2},\frac{q}{4}{\rm e}^{\pi {\rm i}} } \right)
$$
(see, e.g., $(13.2.7)$). The $U$-function may be approximated in terms of Bessel functions as follows:
\begin{align*}
U\!\left( { - n,\frac{1}{2},\frac{q}{4}{\rm e}^{\pi {\rm i}} } \right) & \sim ( - 1)^n {\rm e}^{\pi {\rm i/4}} \left( {\frac{q}{{4n}}} \right)^{1/4} {\rm e}^{ - q/8} n!J_{ - 1/2} ({\rm e}^{\pi {\rm i/2}} \sqrt {nq} ) \\ & = ( - 1)^n \left( {\frac{q}{{4n}}} \right)^{1/4} {\rm e}^{ - q/8} n!I_{ - 1/2} (\sqrt {nq} )
\end{align*}
as $n\to +\infty$ for any fixed positive $q$. Using Stirling's formula and the known large argument asymptotics of the $I$ Bessel function then yields
$$
S(q,n) \sim \frac{{2^{n - 1} }}{{(\pi n)^{1/4} }}\exp \left( {\sqrt {nq}  - \frac{q}{8}} \right)
$$
as $n\to +\infty$ for any fixed positive $q$.
For your related sum the above gives
$$
\sum_{k = 0}^n q^{n-k} \frac{\sqrt{(2n)!}}{(2k)!(n -k)!} \sim \frac{{(2q)^n }}{{2(\pi n)^{1/4} }}\exp\! \bigg( \sqrt {\frac{n}{q}}  - \frac{1}{{8q}} \bigg)
$$
as $n\to +\infty$ for any fixed positive $q$. The treshold is at $q=\frac{1}{2}$ and your sum diverges if $q \ge \frac{1}{2}$. Your numerics with $q=3$ must have some error in it (or you mixed up $3$ and $\frac{1}{3}$).
