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
1+22+33+44+55+
Does it help if I set it equal to x? Or I mean what can I possibly do?
x=1+22+33+44+55+
x2=1+22+33+44+55+
x^2-1=2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}
\frac{x^2-1}{2}=\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}
\left(\frac{x^2-1}{2}\right)^2=2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}
\left(\frac{x^2-1}{2}\right)^2-2=3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}
\vdots

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

Answer

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.

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

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