# What is the value of $\sqrt{1+ \sqrt[3]{1+\sqrt[4]{1+ \sqrt[5]{1+ \cdots }}}}$?

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…

• The value is, numerically, around $1.5176001678777\dots$. The inverse symbolic calculator gives no closed form. – Wojowu Mar 9 '18 at 7:08
• Is this' a well known constant? – Sonal_sqrt Mar 9 '18 at 7:14
• @TheSimpliFire Getting to the 9th root, the value gets above $1.51746$. Up to the 15th root I get $1.51760016787772$ and it doesn't seem to increase any more at least within the accuracy I took. Taking higher accuracy, the number of correct digits seems to grow linearly with however many roots we take. – Wojowu Mar 9 '18 at 8:01
• The first value in the comment above was supposed to be $1.5176$. – Wojowu Mar 9 '18 at 8:08
• @user9198116 See this similar question. – Toby Mak Mar 9 '18 at 12:39

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}

• The question was "Now how can I solve (2)?". But if you're lucky, nobody will notice, and downvote. – Professor Vector Mar 11 '18 at 19:06
• This answer is good because it proves a good upper bound, while truncating the roots provides only a lower bound. – Yuriy S Apr 17 '18 at 20:16
• However, I need to point out that the first three terms of Michael's series aleady give a better upper bound: $$1+\frac{1}{2}+\frac{1}{24}=1.54167$$ – Yuriy S Apr 17 '18 at 20:22

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.
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
$$1+\frac12x+\frac1{24}x^2-\frac5{144}x^3+\cdots$$
• A good idea! The first few terms (Mathematica) look like this: $$1+\frac{x}{2}+\frac{x^2}{24}-\frac{5 x^3}{144}+\frac{637 x^4}{51840}+\frac{143 x^5}{777600}-\frac{9958073 x^6}{2612736000}+\cdots$$ which gives an upper bound of $$L<\frac{590749}{388800}=1.51942 \ldots$$ – Yuriy S Apr 17 '18 at 20:26