How can I prove this limit result associating with an infinite nested radical

Question

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)

Answer 1

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:

1. $x_{k,n}=\sqrt{k+x_{k+1,n}}$
2. 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 $k\in\mathbb{N}$.\
3. 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.