Let $a_n=\sum_{i=0}^n (-2)^i {n \choose i}\frac{(2n-i)!}{(2n)!}$, let $b_n=(1-\frac{1}{n})^n$, and let $c_n=\sum_{i=0}^n (-1)^i \frac{1}{i!}$.
We will show below that $b_n\le a_n\le c_n$ for $n\ge2$, so then we can conclude that
$\displaystyle\lim_{n\to\infty}a_n=\frac{1}{e}$
by the Squeeze Theorem since $\displaystyle\lim_{n\to\infty}b_n=\frac{1}{e}$ and $\displaystyle\lim_{n\to\infty}c_n=\frac{1}{e}$.
$\textbf{1)}$ To show that $a_n\le c_n$ for $n\ge 2$, let $$c_n-a_n=\sum_{i=0}^n (-1)^i \frac{1}{i!}-\sum_{i=0}^n (-2)^i {n \choose i}\frac{(2n-i)!}{(2n)!}$$
$$=\sum_{i=0}^n (-1)^i\frac{1}{i!}\bigg[1-\frac{2^{i}n!}{(n-i)!}\frac{(2n-i)!}{(2n)!}\bigg]=\sum_{i=0}^n(-1)^it_i,$$where $t_i=\frac{1}{i!}\bigg[1-\frac{2^{i}n!}{(n-i)!}\frac{(2n-i)!}{(2n)!}\bigg]>0$ for $i\ge 2$ since
$\frac{2^{i}n!}{(n-i)!}\frac{(2n-i)!}{(2n)!}=\frac{(2n)(2n-2)(2n-4)\cdots(2n-2i+2)}{(2n)(2n-1)(2n-2)\cdots(2n-i+1)}<1$.
Therefore to show that $c_n-a_n\ge0$ for $n\ge2$, it suffices to show that
$t_{i}-t_{i+1}\ge0$ for any even integer $i\ge2$:
$$\displaystyle t_{i}-t_{i+1}=\frac{1}{i!}\bigg[1-\frac{2^{i}n!}{(n-i)!}\frac{(2n-i)!}{(2n)!}\bigg]-\frac{1}{(i+1)!}\bigg[1-\frac{2^{i+1}n!}{(n-i-1)!}\frac{(2n-i-1)!}{(2n)!}\bigg]$$
$$\displaystyle=\frac{1}{(i+1)!}\bigg[i+1-\frac{(i+1)2^{i}n!(2n-i)!}{(n-i)!(2n)!}-1+\frac{2^{i+1}n!(2n-i-1)!}{(n-i-1)!(2n)!}\bigg]$$
$$\displaystyle=\frac{1}{(i+1)!}\bigg[i-\frac{(i+1)2^{i}n!(2n-i)!}{(n-i)!(2n)!}+\frac{2^{i+1}(n-i)n!(2n-i-1)!}{(n-i)!(2n)!}\bigg],$$
so $t_{i}-t_{i+1}\ge0 \iff$ $$ i(n-i)!(2n)!\ge(i+1)2^{i}n!(2n-i)!-2^{i+1}(n-i)n!(2n-i-1)!$$
$$\\\ \;\;\;\;\; =\ 2^{i}n!(2n-i-1)!\big[(i+1)(2n-i)-2(n-i)\big]$$
$$\\\ \;\;=2^{i}n!(2n-i-1)!(i)(2n-i+1)\iff$$
$$(n-i)!(2n)!\ge2^{i}n!(2n-i-1)!(2n-i+1)\iff$$
$$\big[(2n)(2n-1)(2n-2)\cdots(2n-i+2)\big](2n-i)\ge2^{i}n(n-1)(n-2)\cdots(n-i+1)\iff$$
$$\big[(2n)(2n-1)(2n-2)\cdots(2n-i+2)\big](2n-i)\ge(2n)(2n-2)(2n-4)\cdots(2n-2i+2),$$
which is clearly true for any even integer $i\ge2$.
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$\textbf{2)}$ To show that $a_n\ge b_n$ for $n\ge2$, we can use the Binomial Formula and then proceed as above:
Let $$a_n-b_n=\sum_{i=0}^n (-2)^i {n \choose i}\frac{(2n-i)!}{(2n)!}-\sum_{i=0}^n {n\choose i}\big(\frac{-1}{n}\big)^i$$
$$=\sum_{i=0}^n (-1)^{i} \frac{1}{i!}\bigg[\frac{2^{i}n!(2n-i)!}{(n-i)!(2n)!}-\frac{n!}{(n-i)!(n^{i})}\bigg]=\sum_{i=0}^n(-1)^i s_i,$$
where $s_i=\frac{1}{i!}\bigg[\frac{2^{i}n(n-1)\cdots(n-i+1)}{(2n)(2n-1)\cdots(2n-i+1)}-\frac{n(n-1)\cdots(n-i+1)}{n^i}\bigg]>0$ for $i\ge2$ since $\;\;\;\;2^{i}n^{i}=(2n)^{i}\ge(2n)(2n-1)(2n-2)\cdots(2n-i+1).$
Therefore to show that $a_n-b_n\ge0$ for $n\ge2$,
it suffices to show that $s_{i}-s_{i+1}\ge0$ for any even integer $i\ge2$:
$$s_{i}-s_{i+1}=\frac{1}{i!}\bigg[\frac{2^{i}n(n-1)\cdots(n-i+1)}{(2n)(2n-1)\cdots(2n-i+1)}-\frac{n(n-1)\cdots(n-i+1)}{n^i}\bigg]-\frac{1}{(i+1)!}\bigg[\frac{2^{i+1}n(n-1)\cdots(n-i)}{(2n)(2n-1)\cdots(2n-i)}-\frac{n(n-1)\cdots(n-i)}{n^{i+1}}\bigg]$$
$=\frac{1}{(i+1)!}\bigg[\frac{2^{i}(i+1)n(n-1)\cdots(n-i+1)}{(2n)(2n-1)\cdots(2n-i+1)}-\frac{(i+1)n(n-1)\cdots(n-i+1)}{n^i}-\frac{2^{i+1}n(n-1)\cdots(n-i)}{(2n)(2n-1)\cdots(2n-i)}+\frac{n(n-1)\cdots(n-i)}{n^{i+1}}\bigg]$
$=\frac{1}{(i+1)!}\bigg[\frac{2^{i}(i+1)(2n-i)n(n-1)\cdots(n-i+1)}{(2n)(2n-1)\cdots(2n-i)}-\frac{(n-1)(n-2)\cdots(n-i+1)[n(i+1)-(n-i)]}{n^i}\bigg]$,so
$s_i-s_{i+1}\ge0\iff$
$2^{i}n^{i}(i+1)(2n-i)n(n-1)(n-2)\cdots(n-i+1)\ge i(n+1)(n-1)(n-2)\cdots(n-i+1)(2n)(2n-1)\cdots(2n-i)\iff$
$$2^{i}n^{i}(i+1)(2n-i)n\ge i(n+1)(2n)(2n-1)(2n-2)\cdots(2n-i)\iff$$
$$(2n)^{i}(i+1)n\ge i(n+1)(2n)(2n-1)(2n-2)\cdots(2n-i+1),$$
and this inequality is valid since $(i+1)n\ge i(n+1)$ and $(2n)^{i}\ge (2n)(2n-1)(2n-2)\cdots(2n-i+1)$.