How prove this $\lim_{n\to\infty}\sqrt{n}\cdot\sqrt[n]{l-a_{n}}=\frac{\sqrt{e}}{2}$ 
Define the sequence $\{a_{n}\}_{n\ge 2}$ by
  $$a_{n}=\sqrt{1+\sqrt{2+\cdots+\sqrt{n}}}.$$ It is well known
  $a_{n}\longrightarrow l$ for a certain real number $l$. Show that
  $$\lim_{n\to\infty}\sqrt{n}\cdot\sqrt[n]{l-a_{n}}=\dfrac{\sqrt{e}}{2}.$$

This is a nice result, but I can't find a solution. Thank you everyone.
 A: 1) We want to estimate the difference $l-x_n$. To do this note that
$$
l-x_n=\sum\limits_{m\geq n}(x_{m+1}-x_m)
$$
So we find asymptotics for $x_{m+1}-x_m$
2) Show that
$$
x_{n+1}-x_n=\frac{\sqrt{n+1}}{\prod_{1\leq k\leq n}(x_{n+1}^{(k)}+x_n^{(k)})}
$$
where $x_n^{(k)}=\sqrt{k+\sqrt{(k+1)+\ldots+\sqrt{n}}}$. 
3) Using $x_n^{(k)}=\sqrt{k+x_n^{(k+1)}}$ show that
$$\sqrt{k}\leq x_n^{(k)}\leq \sqrt{k}+1\tag{3.1}$$
$$x_n^{(k)}=\sqrt{k}+\frac{1}{2}+\mathcal{O}\left(\frac{1}{\sqrt{k}}\right)\tag{3.2}$$ 
$$x_n^{(k)}=\sqrt{k}\exp\left(\frac{1}{2\sqrt{k}}+\mathcal{O}\left(\frac{1}{k^{3/2}}\right)\right)\tag{3.3}$$
for $1\leq k<n$.
What is more, constants in $\mathcal{O}$ symbols are independent of $k$ and $n$. 
4) Using 3.3 show that 
$$
x_{n+1}-x_n=\frac{1+\mathcal{o}(1)}{2^n\sqrt{(n-1)!}}\exp\left(-\sum\limits_{1\leq k< n}\left(\frac{1}{2\sqrt{k}}+\mathcal{O}\left(\frac{1}{k^{3/2}}\right)\right)\right)
$$
hence
$$
x_{n+1}-x_n\sim\frac{\operatorname{const}}{2^n\sqrt{(n-1)!}}\exp(-\sqrt{n})
$$
and
$$
l-x_n\sim x_{n+1}-x_n\sim \frac{\operatorname{const}}{2^n\sqrt{(n-1)!}}\exp(-\sqrt{n})
$$
5) Applying Stirling's formula we get
$$
\sqrt[n]{l-x_n}\sim\frac{1}{2}\sqrt{\frac{e}{n}}
$$
and the result follows.
