I came across this sum while studying a problem in graph theory, and I'm trying to find a good upper bound. Replacing $(2k)!$ with $(2n)!$ is too coarse for what I need, so I'd like to find some more accurate way to bound the sum. If it helps, here $x = \frac{\log n}{2n^2}$.
I tried Stirling, but I can't figure out how to get anything meaningful out of a sum with $k^k$. I also tried snake oil on $n$, but I still end up with a $(2k)!$ in the sum that I don't know how to deal with.
Are there any identities that might be useful here? Any suggestions would be appreciated. Thanks.
By the Euler integral $$ (2k)! = \int_0^{ + \infty } {e^{ - t} t^{2k} dt} , $$ whence \begin{align*} \sum\limits_{k = 0}^n {\binom{n}{k}(2k)!x^k } & = \sum\limits_{k = 0}^n {\binom{n}{k}\int_0^{ + \infty } {e^{ - t} t^{2k} dt} x^k } = \int_0^{ + \infty } {e^{ - t} \sum\limits_{k = 0}^n {\binom{n}{k}(t^2 x)^k } dt} \\ & = \int_0^{ + \infty } {e^{ - t} (1 + t^2 x)^n dt} . \end{align*} Perhaps this form can be used to obtain better bounds.