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What I've done was assuming this had a solution, therefore

$$ \sqrt[3]{5\sqrt[3]{3\sqrt[3]{5\sqrt[3]{3}}}}\cdots = x $$

By raising both sides of an equation to 3

$$(\sqrt[3]{5\sqrt[3]{3\sqrt[3]{5\sqrt[3]{3}}}}\cdots)^3 = x^3$$ $$5\sqrt[3]{3\sqrt[3]{5\sqrt[3]{3}}}\cdots = x^3$$

By raising both sides of an equation to 3 again, we have that

$$(5\sqrt[3]{3\sqrt[3]{5\sqrt[3]{3}}}\cdots)^3 = x^{3^{3}}$$ $$375\sqrt[3]{5\sqrt[3]{3\sqrt[3]{5\sqrt[3]{3}}}}\cdots = x^{9}$$

But, x was defined as $\sqrt[3]{5\sqrt[3]{3\sqrt[3]{5\sqrt[3]{3}}}}\cdots$, therefore:

$$375x=x^9$$ $$375=x^8$$ $$x=\sqrt[8]{375}$$

After checking the results with a Python program I created, it does seem like the result converges, but I can't find a way to prove it does. I'm a mechanical engineer student so I am not very familiar with proofs and such things.

Rodrigo Bandeira
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  • just draw the graph of $f(x) =(375x)^{1/9},$ call your $L$ the point where that graph crosses the graph $y=x.$ rougly 2.1 . For $0 < x < L$ we see that $ x < f(x) <L.$ Next, say for $ x \geq 1, $ we find $0 < f'(x) < 1/4$ to give a loose bound. – Will Jagy Feb 16 '24 at 04:32
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    Anyway, $x \geq 1,$ by the Mean Value Theorem, $0 < L- f(x) < \frac{1}{4} (L-x)$ – Will Jagy Feb 16 '24 at 04:34
  • See https://math.stackexchange.com/questions/589288/sqrt7-sqrt7-sqrt7-sqrt7-sqrt7-cdots-approximation – lab bhattacharjee Feb 16 '24 at 04:52

3 Answers3

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In order to show something, we need that something clearly defined in our hands. We need to show that an "expression converges". Well, it is a sequence. Then which are its terms? So as the dots are placed in the expression "$\displaystyle \sqrt[3]{5\sqrt[3]{3\sqrt[3]{5\sqrt[3]{3 \cdots}}}}$" it looks like we are starting with the "infinity of the dots", then add some more terms at the beginning. Well, this cannot be defined so. But what we can (recursively) define is the following sequence: $$ \begin{aligned} x_0 &= 1\ ,\\ x_{n+1} &= \sqrt[3]{5\sqrt[3]{3x_n}}\ . \\[3mm] &\qquad\text{ Let $y_n=\ln x_n$ .}\\ &\qquad\text{ (Idea: products become sums.)}\\ &\qquad\text{ The recurrence for $y_n$ is:}\\[3mm] y_0 &=\ln 1 =0\ ,\\ y_{n+1}&=\ln x_{n+1} \\ &=\ln \sqrt[3]{5\sqrt[3]{3x_n}} =\frac 13\ln \left(5\sqrt[3]{3x_n}\right) \\ &=\frac 13\ln 5 +\frac 19 \ln(3x_n) \\ &=\underbrace{\left(\frac 13\ln 5 +\frac 19 \ln 3\right)}_{:=A} + \frac 19 \underbrace{\ln x_n}_{=y_n} \\ &=A+\frac 19y_n\ . \\ &\qquad\text{ So the first terms are:}\\[3mm] x_0 &= 1 & y_0 &= 0\ ,\\ x_1 &= \sqrt[3]{5\sqrt[3]{3}} & y_1 &=\frac 13\ln 5 +\frac 19\ln 3\ ,\\ x_2 &= \sqrt[3]{5\sqrt[3]{3\sqrt[3]{5\sqrt[3]{3}}}} & y_2 &=\frac 13\ln 5 +\frac 19\ln 3 + \frac 1{27}\ln 5 +\frac 1{81}\ln 3\ ,\\ \end{aligned} $$ and so on. It is easier to work with $y_n$, the "additive" version of $x_n$, and the convergence is to $$ \begin{aligned} \lim y_n &= \frac 13\ln 5\left(1+\frac 19+\frac 1{81}+\dots\right) + \frac 19\ln 3\left(1+\frac 19+\frac 1{81}+\dots\right) \\ &= \underbrace{ \frac 19(3\ln 5+\ln 3)}_{=A}\cdot \frac 1{1-\frac 19} \\ &= \frac 18(3\ln 5+\ln 3) =\frac 18\ln (3\cdot 5^3)\ .\qquad\text{ So:} \\ \lim x_n &= \lim \exp y_n \\ &=\exp\lim y_n\\ &=(3\cdot 5^3)^{1/8}=375^{1/8}\approx 2.09775086317869667712859368405659872\dots\ . \end{aligned} $$

In essence we have in the $y$-world a linear combination of (the same) geometric series.

LATER EDIT: ($A$ was introduced above. The answer is no longer so short and rather a hint, but it allows to illustrate the idea and the convergence.) The first terms are explicitly

  • $y_0=0$,
  • $y_1=A+\frac 19y_0=\frac 19$,
  • $y_2=A+\frac 19y_1=A+\frac 19A$,
  • $y_3=A+\frac 19\left(A+\frac 19A\right)=A+\frac 19 A+\frac 1{9^2}A$,
  • $y_4=A+\frac 19\left(A+\frac 19A+\frac 1{9^2}A\right)=A+\frac 19 A+\frac 1{9^2}A+\frac 1{9^3}A$,

and so on. The "obvious formula" for $y_n$ can be easily shown inductively. In it, the convergence and the limit are transparent.

dan_fulea
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  • Thanks for the great answer. I understood it pretty well, except the part where you just got this yn = ln (xn). Why does the sum of yn equals to the original sequence? – Rodrigo Bandeira Feb 16 '24 at 06:05
  • @JyrkiLahtonen The answer is now edited, indeed a good suggestion! Your comments are always pointed and helpful. Always welcome! – dan_fulea Feb 16 '24 at 12:59
  • @RodrigoBandeira The idea to pass from the $x_n$, with an expression involving products and radicals of order $3,5$, to $y_n$, with and expression involving in parallel sums and multiplication with $\frac 13$, $\frac 15$, via$$\ln$$ is natural. First of all it is no longer a painful typing. (And when typing is cumbersome, it is a sign that the structure is not in its suitable structural shape...) Then the operations involved are (psychologically) easier. In this $y$-world the recursion, the formula for the general term, the convergence and the limit are clear. Now we use $$\exp$$ to come back. – dan_fulea Feb 16 '24 at 13:04
  • Good job! ${}{}$ – Jyrki Lahtonen Feb 16 '24 at 13:42
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From $x_0 = 5 \sqrt[3]{3}$, we repeatedly do $y_{i + 1} = 3\sqrt[3]{y_i}$ and $x_{i+1} = 5 \sqrt[3]{y_{i+1}}$, i.e., $x_{i+1} = 5 \sqrt[3]{3} \sqrt[9]{x_i}$.

We can see that for positive $x$, (1) $x_{i+1} > x_{i}$ when $x_i < 5\sqrt[8]{135}$, and (2) when $x_i < 5\sqrt[8]{135}$, $x_i < x_{i+1} < 5\sqrt[8]{135}$. Also $x_0 = 5 \sqrt[3]{3} < 5\sqrt[8]{135}$.

Then we conclude that the sequence is convergent by the monotone convergence theorem.

Vezen BU
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Firstly - thanks for posting as I've enjoyed working through this.

Okay, let your number be $S$ with

\begin{equation} S = \sqrt[3]{5\sqrt[3]{3\sqrt[3]{5\sqrt[3]{3 \cdots}}}} \end{equation}

We notice that $S$ repeats itself within itself, i.e. \begin{equation} S = \sqrt[3]{5\sqrt[3]{3S}} \end{equation} And so, \begin{equation} S^3 = 5\sqrt[3]{3S} \longrightarrow S^9 = 5^3\cdot 3S --> S^9 - 5^3\cdot 3S = 0 \end{equation} From here, we factor our equation: \begin{align} &0 = S\left(S^8 - 5^3\cdot 3\right) \nonumber \\ &0 =S\left(S^4 - \sqrt{5^3\cdot 3}\right) \left(S^4 + \sqrt{5^3\cdot 3}\right) \\ &0 = S\left(S^2 - \sqrt{\sqrt{5^3\cdot 3}}\right) \left(S^2 + \sqrt{\sqrt{5^3\cdot 3}}\right) \left(S^4 + \sqrt{5^3\cdot 3}\right) \\ &0 = S\left(S - \sqrt{\sqrt{\sqrt{5^3\cdot 3}}}\right) \left(S + \sqrt{\sqrt{\sqrt{5^3\cdot 3}}}\right) \left(S^2 + \sqrt{\sqrt{5^3\cdot 3}}\right) \left(S^4 + \sqrt{5^3\cdot 3}\right) \end{align} Given we know that $S$ is both Real and Positive, we find the only solution is \begin{equation} S - \sqrt{\sqrt{\sqrt{5^3\cdot 3}}} = 0 \longrightarrow S = \sqrt{\sqrt{\sqrt{5^3\cdot 3}}}= 5^{\frac{3}{8}}3^{\frac{1}{8}} \end{equation}

David Galea
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