How can we find the value of $$\sqrt{1+ \sqrt[3]{1+\sqrt[4]{1+ \sqrt[5]{1+ \cdots}}}}=?$$
My Approach:
Let $$f(n)=\sqrt[n]{1+ \sqrt[n+1]{1+\sqrt[n+2]{1+ \sqrt[n+3]{1+ \cdots}}}} \tag{1},$$
then $f(2)$ is our solution.
So, doing $n$th power in both sides of $(1)$, we get: $${ \{ f(n) \} }^n =1+f(n+1)$$ $$\implies { \{ f(n) \} }^n - f(n+1) = 1 \tag{2}$$ Now how can I solve $(2)$ ? Any help please…
On
Following on from TheSimplifier, you can try to evaluate $$\left(1+x\left(1+x\left(\cdots\right)^{1/4}\right)^{1/3}\right)^{1/2}$$ at the value $x=1$.
It has the advantage of a finite calculation for each coefficient of $x^n$. But I think the first few terms are
$$1+\frac12x+\frac1{24}x^2-\frac5{144}x^3+\cdots$$
On
Estimation of upper bound.
L signs the value of the expression, then we can write:
\begin{align}L^2=1+\left(1+(1+\cdots+(1+(1+s)^{{1/n}})^{{1/{n-1}}}\cdots)^{1/4}\right)^{1/3}\end{align}
where
\begin{align}s=\sqrt[n+1]{1+\sqrt[n+2]{1+ \sqrt[n+3]{1+ \cdots}}} \tag{1}\end{align}
Accordance with Bernoulli$^,$s inequality if $0\le{r}\le1$ and $x\ge-1$ then $(1+x)^r\le(1+xr)$
In our case $r=\frac{1}{k}, k=3,4, ....n $ and $s\gt1 $.
First apply to $(1+s)^{1/n}$ we have that $\le1+\frac{s}{n}$ then outward apply to all power factor we get the following:
\begin{align}L^2\le1+1+\left(1+(1+\cdots+(1+(1+\frac{s}{n}))\frac{1}{n-1}\cdots)\frac{1}{4}\right)\frac{1}{3}\end{align}
Perform the multiplications:
\begin{align}L^2\le2 +2(\frac{1}{3!} +\frac{1}{4!}+......\frac{1}{(n-1)!}+\frac{s}{n!})\end{align}
Namely
\begin{align}L^2\le2 +2\sum_{k=3}^n\frac{1}{k!}\end{align}
We could apply to $s$ the above procedure then we have:
\begin{align}L^2\le2 +2\sum_{k=0}^{\infty}\frac{1}{k!}-5\end{align}
\begin{align}L\le \sqrt{2e-3}=1,56094\end{align}
Comment, not an answer
We have $$\begin{align}f(2)=\sqrt{1+ \sqrt[3]{1+\sqrt[4]{1+ \sqrt[5]{1+ \cdots}}}}=\left(1+\left(1+(1+\cdots)^{1/4}\right)^{1/3}\right)^{1/2}\end{align}$$ From the Binomial Theorem, $$\begin{align}(1+x)^{1/n}&=1+\frac1nx-\frac{n-1}{n\cdot2n}x^2+\frac{(n-1)(2n-1)}{n\cdot2n\cdot3n}x^3-\cdots\\&=1+\sum_{m=1}^\infty \left(x^m\prod_{k=0}^m\frac{(-1)^{k+1}(kn-1)}{(k+1)n}\right)\end{align}$$ so $$f(2)=1+\sum_{m=1}^\infty \left(\left(1+\sum_{m=1}^\infty \left(\left(\cdots\right)^m\prod_{k=0}^m\frac{(-1)^{k+1}(3k-1)}{3(k+1)}\right)\right)^m\prod_{k=0}^m\frac{(-1)^{k+1}(2k-1)}{2(k+1)}\right)$$ which is essentially an infinite nest of iterations of simple summations and products.
There may be techniques to evaluate this, but none of which I am aware of.