What is the general method to find an asymptotic formula for alternating sums 
How to find an asymptotic formula for the following sum:
  $$ \sum_{k=0}^n(-1)^k{n\choose k}\frac{n!2^k}{(n-k)!}(2n-2k)! $$ when $n\to\infty$?


If we set $S_k:={n\choose k}\frac{n!2^k}{(n-k)!}(2n-2k)!$, then we know that $S_k$ is a nonnegative decreasing sequence with regard to $k$. I have tried to bound the sum by the first three terms as upper bound and the first four terms as the lower bound but still cannot get an asymptotic formula. I mean this series goes very slow when $k$ grows larger.
And this formula is deduced from the inclusion-exclusion principle and that's why we have the alternating $1, -1$.
So in general, I am curious how to find an asymptotic formula when we use the inclusion-exclusion formula to get the explicit formula.
I have referred to this MO post but nothing in that post works to me.

Remark: Also, please give me a hint on the problem or you may describe the methods generally, please leave me to finish the details. Thank you! 
 A: $$
\begin{align}
\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{n!2^k}{(n-k)!}(2n-2k)!
&=n!^2\sum_{k=0}^n(-1)^k\frac{2^k}{k!}\binom{2n-2k}{n-k}\tag1\\
&\sim n!^2\sum_{k=0}^n(-1)^k\frac{2^k}{k!}\frac{4^{n-k}}{\sqrt{\pi\left(n-k+\frac14\right)}}\tag2\\
&=\frac{4^nn!^2}{\sqrt{\pi n}}\sum_{k=0}^n(-1)^k\frac{2^{-k}}{k!\sqrt{1-\frac kn+\frac1{4n}}}\tag3\\
&\sim\frac{4^nn!^2}{\sqrt{\pi n}}\sum_{k=0}^n(-1)^k\frac{2^{-k}}{k!}\left(1+\frac{k}{2n}-\frac1{8n}\right)\tag4\\
&\sim\frac{4^nn!^2}{\sqrt{\pi ne}}\left(1-\frac3{8n}\right)\tag5
\end{align}
$$
Explanation:
$(1)$: rearrange the factorials
$(2)$: apply $(10)$ from this answer
$(3)$: move factors around
$(4)$: $(1-x)^{-1/2}\sim1+x/2$
$(5)$: apply the power series for $e^x$
A: Using Stirling's approximation, we get that
$$\sum_k (-1)^k S_k \approx -(n!)^2\sum_{k=1}^{n-1} (-2)^k\frac{\sqrt{2n-2k}}{\sqrt{k}(n-k)} \frac{(2n-2k)^{(2n-2k)}}{k^k(n-k)^{(2n-2k)}}e^k$$
$$ = -2^{2n+\frac{1}{2}}(n!)^2\sum_{k=1}^{n-1} \left(-\frac{e}{2k}\right)^k\frac{1}{\sqrt{k(n-k)}} \approx -\frac{2^{2n+\frac{1}{2}}(n!)^2}{\sqrt{n}}\sum_{k=1}^{N}\left(-\frac{e}{2k}\right)^k\frac{1}{\sqrt{k}}$$
where we consider the contribution from the terms past $N$, a large enough integer hopefully much much smaller than $n$, to be negligible. The terms of the series decay rapidly, so the approximation $n-k \approx n$ is valid. We could collect the first few terms to get a semi analytic, but messy, coefficient but the sum is approximately $1.08$. So we get
$$\sum_{k=0}^n (-1)^k S_k \approx 2.16 \pi\sqrt{2n} \left(\frac{2n}{e}\right)^{2n}$$
by Stirling's approximation again.
