$$a_n=\sqrt{1+\sqrt{2+\cdots\sqrt{n}}}$$
We can prove that $\{a_n\}$ is convergent
(using mathematical induction, $\sqrt {k+\sqrt{k+1+\cdots\sqrt{n}}}\leq k-1, for \ k\geq3$).
If $$ \lim\limits_{n\to\infty} a_n=\ell, $$ prove:
$$\lim\limits_{n\to\infty} \left[\,(\ell-a_n)^{1/n}\cdot n^{1/2}\,\right]=\frac{\sqrt e}{2}$$
Let's denote for $x_i\ge 0$: $$ [x_1, x_2,\cdots ,x_n] = \sqrt{x_1 + \sqrt{x_2 + \cdots \sqrt{x_n}}}$$
Let $$c_{k,n} = [k, k+1, \cdots, n] \text{, for } n \ge k$$ from your inequality proved by induction we deduce:
$$c_{k,n} = [k + c_{k+1,n}] \le \sqrt{2k} \text{, for } n \ge k \ge 3$$
And injecting this enquality again, we deduce $$\sqrt{k} \le c_{k,n} \le \sqrt{k + \sqrt{2(k+1)} }$$
We have in particular $\ell = c_{1,\infty}$
From $$c_{k,\infty} - c_{k,n} = \frac{c_{k+1, \infty} - c{k+1, n}}{c_{k,\infty} + c_{k,n}}$$
we deduce
$$\ell -a_n = \frac{c_{n+1,\infty}}{\prod_{k=1}^{n}\left(c_{k,\infty} + c_{k, n}\right)}$$
Let be $\varepsilon > 0$, there exists $N_0$ large enough, such that
$$\sqrt{k} < \sqrt{k+\sqrt{2(k+1)}} < (1 + \varepsilon)\sqrt{k}$$
Let be $K_{N_0} = \frac{\sqrt{N_0!}}{\prod_{k=1}^{N_0}c_{k,\infty}}$ So for all $n\ge N_0$, we have
$$ K_{N_0} \frac{ \sqrt{n+1} }{ (1+\varepsilon)^{n - N_0} 2^n \sqrt{n!} } \le K_{N_0}\frac{ c_{n+1,\infty} }{ (1+\varepsilon)^{n - N_0} 2^n \sqrt{n!}} \le \ell - a_n \le \frac{ c_{n+1,\infty} }{ 2^n \sqrt{n!} } \le \frac{ \sqrt{2(n+1)} }{ 2^n\sqrt{n!} }$$
Hence by taking the nth square root, we deduce
$$ K_{N_0}^{\frac1{n}} \frac{\left(n+1\right)^{\frac1{2n}}}{2\left(1+\varepsilon\right)^{1 - n/N_0}\left(n!\right)^{\frac1{2n}}} \le \left(\ell - a_n\right)^{\frac1{n}} \le \frac{(2(n+1))^{\frac1{2n}}}{2\left(n!\right)^{\frac1{2n}}}$$ We have then by Sterling formula
There exists $N_1 \ge N_0$ s.t. for all $n \ge N_1$, we have
$$ (1 - 2\varepsilon)\frac{\sqrt{e}}{2} \le \sqrt{n}\left(\ell - a_n\right)^{\frac1{n}} \le (1 + 2\varepsilon)\frac{\sqrt{e}}{2}$$
