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}}}}}

Is it possible to find the value of
Does it help if I set it equal to x? Or I mean what can I possibly do?

I don’t see it’s going anywhere. Help appreciated!


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

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}}}},

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.

Source : Link , Question Author : user84940 , Answer Author : Antonio Vargas

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