Whether $\lim_{n\to \infty} \frac{2}{\mathsf{e}}(\sum_{k=0}^{\lfloor n/2\rfloor} \binom{n}{k}(1-\frac{2k}{n})^{n-1})^{-1/n}$ exists 
Problem: Decide whether or not
$\lim_{n\to \infty} \frac{2}{\mathsf{e}}\left(\sum_{k=0}^{\lfloor n/2\rfloor} \binom{n}{k} \left(1-\frac{2k}{n}\right)^{n-1}\right)^{-1/n}$ exists.

Background Information: I encountered this problem, when I tried to answer the following question
Interval of convergence of Lagrange's infinite series
In my answer there, I $\color{blue}{\textrm{GUESS}}$ that the limit exists and equals to the Laplace limit $0.66274 34193 49181 58097 47420 97109 25290...$
which is the solution of the equation $x \mathrm{e}^{\sqrt{1+x^2}} = 1 + \sqrt{1+x^2}$.
(For Laplace limit and more information, see Ref. [1]-[4].)
Let $B_n \triangleq \frac{2}{\mathsf{e}}\left(\sum_{k=0}^{\lfloor n/2\rfloor} \binom{n}{k} \left(1-\frac{2k}{n}\right)^{n-1}\right)^{-1/n}$.
Some numerical experiments show that $B_n$ is non-increasing.
I tried to prove it, but have not yet succeeded.
When $n=1000$ (Maple can not easily evaluate $B_n$ for larger $n$), $B_{1000} = 0.6627434531...$
Any comments and solutions are welcome.
Reference
[1] https://en.wikipedia.org/wiki/Laplace_limit
[2] http://www.mygeodesy.id.au/documents/Solutions%20of%20Keplers%20Equation.pdf
[3] https://arxiv.org/pdf/1305.3438.pdf
[4] "Orbital Mechanics for Engineering Students", http://www.nssc.ac.cn/wxzygx/weixin/201607/P020160718380095698873.pdf
 A: 0. We write $f(\mathrm{x}) \asymp g(\mathrm{x})$ on $\mathcal{D}$ if there exist constants $c_1, c_2 > 0$ such that $c_1 \leq \frac{f(\mathrm{x})}{g(\mathrm{x})} \leq c_2 $ for all $\mathrm{x} \in \mathcal{D}$.
1. From the Stirling's approximation, we know that
$$ n! \asymp n^{n+\frac{1}{2}}\mathrm{e}^{-n} \quad \text{for} \quad n \geq 1. $$
So, if $n \geq 4$ and $0 < k < \frac{n}{2}$, then by writing $t = \frac{1}{2}-\frac{k}{n}$ (or equivalently $k = \frac{n}{2} - nt$),
\begin{align*}
\binom{n}{k} \left(1-\frac{2k}{n}\right)^{n-1}
&\asymp \frac{n^{n+\frac{1}{2}}}{k^{k+\frac{1}{2}}(n-k)^{n-k+\frac{1}{2}}} \left(1-\frac{2k}{n}\right)^{n-1} \\
&= \frac{n^{n+\frac{1}{2}}}{n^{n+1} \left(\frac{1}{2} - t\right)^{\frac{n}{2}-nt+\frac{1}{2}} \left(\frac{1}{2} + t\right)^{\frac{n}{2}+nt+\frac{1}{2}}} (2t)^{n-1} \\
&= \frac{1}{t\sqrt{n(1-4t^2)}} f(t)^n,
\end{align*}
where $ f : [0, \frac{1}{2}] \to [0, \infty)$ is defined by
$$ f(t) := \frac{2t}{\left(\frac{1}{2} - t\right)^{\frac{1}{2}-t} \left(\frac{1}{2} + t\right)^{\frac{1}{2}+t}}, \qquad f(\tfrac{1}{2}) := \lim_{t \to (1/2)^-} f(t) = 1. $$
Since $\frac{1}{n} \leq \frac{k}{n} \leq \frac{1}{2}-\frac{1}{2n}$, it follows that $\frac{1}{2n} \leq t \leq \frac{1}{2} - \frac{1}{n}$. Using this, it is easy to check that
$$ \log\Biggl(\frac{1}{t\sqrt{n(1-4t^2)}}\Biggr) \asymp \log n, $$
and so, a standard argument shows that
$$ B_n \sim \frac{2}{\mathrm{e}} \Biggl( \sum_{0 \leq k < \frac{n}{2}} f\left(\frac{1}{2}-\frac{k}{n}\right)^n \Biggr)^{-\frac{1}{n}} \xrightarrow[n\to\infty]{} \frac{2}{\mathrm{e}}\left( \max_{0 \leq t \leq \frac{1}{2}} f(t) \right)^{-1}. $$
2. To identify the limit, note that the logarithmic derivative
$$ (\log f(t))' = \frac{1}{t} + \log\left(\frac{1}{2}-t\right) - \log\left(\frac{1}{2} + t\right) $$
is strictly decreasing with $\lim_{t \to 0^+} (\log f(t))' = +\infty$ and $\lim_{t \to (1/2)^-} (\log f(t))' = -\infty$. So the function $f$ achieves a unique maximum point, which corresponds to the unique solution of the equation $(\log f(t))' = 0$ on $(0, \frac{1}{2})$. We denote that maximum point by $\ell$. Then
$$ \lim_{n\to\infty} B_n = \frac{2}{\mathrm{e}f(\ell)} = \frac{\sqrt{1-4\ell^2}}{2\ell}. $$
So if we write $L = \lim_{n\to\infty} B_n$, then $\ell = \frac{1}{2\sqrt{L^2 + 1}}$, and using this, it is not hard to check that
$$ L\mathrm{e}^{\sqrt{L^2+1}} = 1+\sqrt{L^2+1} $$
as desired.
