How can I prove this limit result associating with an infinite nested radical Let $a_n=\sqrt{1+\sqrt{2+\sqrt{3+...+\sqrt{n}}}}$
I can show that $\lim\limits_{n\to∞}a_n$ converges,let $l=\lim\limits_{n\to∞}a_n$
Now what puzzles me is that how to prove $\lim\limits_{n\to∞}\sqrt{n}\sqrt[n]{l-a_n}=\frac{\sqrt{e}}{2}$
I have already figured out that $\lim\limits_{n\to∞}\sqrt{n}\sqrt[n]{l-a_n}\le\frac{\sqrt{e}}{2}$
How about the other half?
P.S. Here's how I work the upper bound out:
Since $a_n$ converges, we have
$\begin{aligned}
l-a_n=& \sum\limits_{i=n}^∞(a_{i+1}-a_i)\\
=& \sum\limits_{i=n}^∞(\sqrt{1+\sqrt{2+\sqrt{3+...+\sqrt{i+1}}}}-\sqrt{1+\sqrt{2+\sqrt{3+...+\sqrt{i}}}})\\
=&\sum\limits_{i=n}^∞\frac{\sqrt{2+\sqrt{3+...+\sqrt{i+1}}}-\sqrt{2+\sqrt{3+...+\sqrt{i}}}}{\sqrt{1+\sqrt{2+\sqrt{3+...+\sqrt{i+1}}}}+\sqrt{1+\sqrt{2+\sqrt{3+...+\sqrt{i}}}}}\\
\le&\sum\limits_{i=n}^∞\frac{\sqrt{2+\sqrt{3+...+\sqrt{i+1}}}-\sqrt{2+\sqrt{3+...+\sqrt{i}}}}{2\sqrt{1}}\\
\le&...(\text{repeat the process n times})\\
\le&\sum\limits_{i=n}^∞\frac{\sqrt{i+1}}{2^i\sqrt{i!}}\\
<&\sum_{i=n}^\infty \frac{\sqrt{i+1}}{2^i\sqrt{i!}}\\
\le&\frac{1}{2^n\sqrt{n!}}\sum_{i=n}^\infty\frac{\sqrt{i+1}}{2^{i-n}}\\
=&\frac{1}{2^n\sqrt{n!}}\sum_{i=0}^\infty\frac{\sqrt{n+i+1}}{2^i}
=\frac{\sqrt{n}}{2^n\sqrt{n!}}\sum_{i=0}^\infty\frac{1}{2^i}+\frac{1}{2^n\sqrt{n!}}\sum_{i=0}^\infty\frac{\sqrt{n+i+1}-\sqrt{n}}{2^i}\\
=&\frac{2\sqrt{n}}{2^n\sqrt{n!}}+\frac{1}{2^n\sqrt{n!}}\sum_{i=0}^\infty\frac{1}{2^i}\cdot\frac{i+1}{\sqrt{n+i+1}+\sqrt{n}}<\frac{2\sqrt{n}+C}{2^n\sqrt{n!}}<\frac{2\sqrt{n}+n}{2^n\sqrt{n!}}
\end{aligned}
$
Now by this estimation of the upper bound, we can show $\lim\limits_{n\to∞}\sqrt{n}\sqrt[n]{l-a_n}\le\frac{\sqrt{e}}{2}$ by calculation(with Stirling's approximation)
 A: This is more like long observation with I think it fits better in this section than in the comment section.
It seems that your bounding came a little too soon (as comedians would say).
For each $n$ and $1\leq k\leq n$ set
\begin{align}
x_n&:=\sqrt{1+\sqrt{2+\ldots +\sqrt{n}}}\\
x_{k,n}&:=\sqrt{k+\sqrt{(k+1)+\ldots +\sqrt{n}}}
\end{align}
The sequence $x_n$ is monotone increasing and bounded (with a little effort, $x_n<2$).
Taking conjugates one gets
\begin{align}
x_{n+1}-x_n &=\sqrt{1+\sqrt{2+\ldots +\sqrt{n+\sqrt{n+1}}}}-\sqrt{1+\sqrt{2+\ldots +\sqrt{n}}}\\
&=\frac{\sqrt{2+\sqrt{3+\ldots +\sqrt{n+\sqrt{n+1}}}}-\sqrt{2+\sqrt{3+\ldots +\sqrt{n}}}}{x_{1,n+1}+x_{1,n}}\\
&=\frac{\sqrt{3+\ldots +\sqrt{n+\sqrt{n+1}}} - \sqrt{3+\ldots +\sqrt{n}}}{(x_{1,n+1}+x_{1,n})(x_{2,n+1}+x_{2,n})}\\
&=\ldots\\
&=\frac{\sqrt{n+1}}{\prod^n_{k=1}(x_{k,n+1}+x_{k,n})}
\end{align}
The key part then is finding good asymptotics for the product $\prod^n_{k=1}(x_{k,n+1}+x_{k,n})$.
We have the obvious relations:

*

*$x_{k,n}=\sqrt{k+x_{k+1,n}}$

*and by induction $\sqrt{k}\leq x_{k,n}<\sqrt{k}+1$:
The right-hand side is obvious; as for the second, notice that $x_{n, n}<\sqrt{n}+1$. Suppose statement holds for all $k+1,k+2,\ldots, n$. Then
\begin{align}
x_{k,n}=\sqrt{k+x_{k+1,n}}\leq \sqrt{k+\sqrt{k+1}+1}\leq \sqrt{k}+1
\end{align}
since $\sqrt{k+1}\leq 2\sqrt{k}$ for all $k\in\mathbb{N}$.\

*Let $\phi(x)=\sqrt{1+x}$. Expanding around $0$ we have
$$\phi(1+x)=\phi(0)+\phi'(0)x+\frac12\phi''(\xi)x^2=1+\frac12 x - \frac14(1+\xi)^{-3/2}x^2$$
where $\xi$ is between $0$ and $x$.
Hence, by (3)
\begin{align}x_{k, n}&=\sqrt{k}\sqrt{1+\tfrac{x_{k+1,n}}{k}}=\sqrt{k}\Big(1+\frac12\frac{x_{k+1,n}}{k}-\frac{1}{4}\frac{1}{(1+\xi_{k,n})^{3/2}}\frac{x^2_{k+1,n}}{k^2}\Big)\\
&=\sqrt{k}+\frac12+O\big(\tfrac{1}{\sqrt{k}}\big)
\end{align}
I think this asymptote is not enough yet to solve the OP.  But I think that is a good start.

I will leave to his as Community wiki in the hope that others, including the OP,  may be able to contribute and have a complete and clear solution.
