Infinitely nested radical question

Question

Here is the problem: $\sqrt{1 + 2\sqrt{3+4\sqrt{5+\cdots}}}$ I have been working on this problem for almost a week but I don't seem to find anything valuable to solve this. I found a pattern that goes like this : $\sqrt{1+2\sqrt{1+2(1+2\sqrt{2+3(1+2\sqrt{3+\cdots}}}}$ But it seems like it is useless. And evaluated this equation too: $F(x)^2 = x+(x+1)F(x+2)$ But again this seems useless too. I am well too confused.

Answer 1

This nested radical should be interpreted as limit of $\sqrt{1+2\sqrt{3+4\sqrt{5+\ldots+2n\sqrt{2n+1}}}}$ or more formally we let $$\tag1 f(k,0)=0$$ $$\tag2 f(k,m+1)=\sqrt{k+(k+1)f(k+2,m)}$$ $$\tag3 F(k)=\lim_{m\to\infty}f(k,m)$$ and are interested in $F(1)$. To show existence of the limit, first note that $$f(k,m)\le f(k,m+1)< k+4,$$ where the first inequality should be clear (or follows readily by induction on $m$) and the second inequality follows easily by induction on $m$ with the crucial step being $$ \begin{align}f(k,m+1)&=\sqrt{k+(k+1)f(k+2,m)}\\&<\sqrt{k+(k+1)(k+6)}\\&=\sqrt{(k+4)^2-10}.\end{align}$$ Therefore, in $(3)$ we take the limit of an increasing and bounded sequence. We conclude that the limit exists and $$F(k)\le k+4.$$ Knowing that the limit exists, we are now allowed to conclude $$ F(k)=\sqrt{k+(k+1)F(k+2)}$$ or $$\tag4F(k)^2=k+(k+1)F(k+2).$$ Since (quite clearly) $F(k+2)> F(k)$, we see that $F(k)$ is larger than the positive root of $p(X)=X^2-(k+1)X-k$, in particular $ F(k)>k+1$.

Letting $$\delta(k)=k+4-F(k)\in[0,3) $$ we obtain from $(4)$ $$\tag52(k+4)\delta(k)-\delta(k)^2=10+(k+1)\delta(k+2).$$ Using the bound $0\le \delta(k)<3$ this gives us $$\tag6\delta(k)>\frac 5{k+4} $$ and $$ (2k+5)\delta(k)<10+(k+1)\delta(k+2),$$ $$\tag7 2\delta(k)+\frac{3\delta(k)-10}{k+1}<\delta(k+2)$$ Assume that for some $k$ we have $\delta(k)> \frac{10}{k+4}$. Then $\frac{3\delta(k)-10}{k+1}>-\delta(k)0$ so that $q:=\frac{\delta(k+2)}{\delta(k)}>1$. But then $\frac{3\delta(k+2)-10}{k+3}>\frac{3\delta(k)-10}{k+1}$ so that $\frac{\delta(k+4)}{\delta(k+2)}\ge q$. By induction, $\delta(k+2m)\ge q^m\delta(k)$, which is absurd. We conclude $$ \delta(k)<\frac{10}{k+4}.$$ In other words, $$F(k)=k+4+\frac{5+\delta_2(k)}{k+4} $$ with $0<\delta_2(k)<5$.

This allows us to compute $F(1)$ with acceptable precision from just a few iterations of $(4)$. Unfortunately, the computed value $$F(1)=3.598544540688569092598791775885364010757216864766804489021\ldots $$ is not recognized by Plouffe's converter.