I would like to calculate this limit. I tried to look for known limits which can be substituted to parts of this limit but couldn't find information about the double factorial.
$$\lim\limits_{k \to \infty}(1+2^{k+1})^{(2^{k-2})-2}\cdot\frac{(2^k-1)^2}{(2^k-1)!!}$$
Thanks ahead for any help
EDIT
Denote $n=2^k-1=2\cdot 2^{k-1}-1$ so we actually need to calculate:
$$\lim\limits_{n \to \infty}(2n+3)^{\frac{n+1}{4}-2}\cdot\frac{n^2}{n!!}$$ And have by a relation-to-factorial formula for odd n:
$$n!! = \frac{2^k!}{2^{2^{k-1}}\cdot 2^{k-1}!} = \frac{(n+1)!}{2^{\frac{n+1}{2}}\cdot (\frac{n+1}{2})!}$$
Now using the Stirling approximation we have (only for odd n):
$$n!! \sim \frac{\sqrt{2\pi(n+1)}(\frac{n+1}{e})^{n+1}}{2^{\frac{n+1}{2}}\cdot \sqrt{2\pi(\frac{n+1}{2})}(\frac{\frac{n+1}{2}}{e})^{\frac{n+1}{2}}} = \frac{\sqrt{2}(\frac{n+1}{e})^{n+1}}{2^{\frac{n+1}{2}}(\frac{n+1}{e})^{\frac{n+1}{2}}(\frac{1}{2})^{\frac{n+1}{2}}} = \sqrt{2}(\frac{n+1}{e})^{\frac{n+1}{2}}$$
From here it is clear to see that for odd n's $$\lim\limits_{n \to \infty}(2n+3)^{\frac{n+1}{4}-2}\cdot\frac{n^2}{n!!} = 0$$
HINT
Let indicate $n=2^k-1\to \infty$ then we have
$$(1+2^{k+1})^{(2^{k-2})-2}\cdot\frac{(2^k-1)^2}{(2^k-1)!!}=(2n+3)^{\frac{n+1}4-2}\cdot\frac{n^2}{n!!}=(2n+3)^{\frac{n-7}4}\cdot\frac{n^2}{2^nn!}$$
then use Stirling approximation or by Root test
$$\sqrt[n]{(2n+3)^{\frac{n-7}4}\cdot\frac{n^2}{2^nn!}} \sim \sqrt[n]{\frac{n^\frac{n}4n^2}{2^{\frac{3n}4}n!}}=\frac{\sqrt[n]{n^2}}{2^\frac34}\cdot\frac{\sqrt[4] n}{\sqrt[n]{n!}}$$