Finding the value of √1+2√2+3√3+4√4+5√5+…\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}

This is meant to follow up on Ethan’s comment about using Herschfeld’s theorem to prove that the expression converges.

Theorem (Herschfeld, 1935). The sequence
$$u_n = \sqrt{a_1 + \sqrt{a_2 + \cdots + \sqrt{a_n}}}$$
converges if and only if
$$\limsup_{n\to\infty} a_n^{2^{-n}} < \infty.$$

The American Mathematical Monthly, Vol. 42, No. 7 (Aug-Sep 1935), 419-429.

In our case we have

$$\begin{align} u_1 &= \sqrt{1}, \\ u_2 &= \sqrt{1 + 2\sqrt{2}} = \sqrt{1 + \sqrt{2^3}}, \\ u_3 &= \sqrt{1 + 2\sqrt{2 + 3\sqrt{3}}} = \sqrt{1 + \sqrt{2^3 + \sqrt{2^4 3^3}}}, \\ u_4 &= \sqrt{1 + \sqrt{2^3 + \sqrt{2^4 3^3 + \sqrt{2^8 3^4 4^3}}}}, \end{align}$$

and so on, so that

$$a_n = n^3 \prod_{k=2}^{n-1} k^{2^{n-k+1}}.$$

We then have

$$a_n^{2^{-n}} = n^{3\cdot 2^{-n}} \prod_{k=2}^{n-1} k^{1/2^{k-1}} \sim \prod_{k=2}^{\infty} k^{1/2^{k-1}}$$

as $n \to \infty$, where the infinite product converges because $k^{1/2^{k-1}} = 1 + O(\log k/2^k)$ as $k \to \infty$. Therefore $u_n$ converges by Herschfeld's theorem.