How to check the convergence of $\sqrt[3]{5\sqrt[3]{3\sqrt[3]{5\sqrt[3]{3}}}}\cdots$

Question

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.

Answer 1

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 $$ \begin{aligned} \lim y_n &= \frac 13\ln 5\left(1+\frac 19+\frac 1{81}+\dots\right) + \frac 19\ln 3\left(1+\frac 19+\frac 1{81}+\dots\right) \\ &= \underbrace{ \frac 19(3\ln 5+\ln 3)}_{=A}\cdot \frac 1{1-\frac 19} \\ &= \frac 18(3\ln 5+\ln 3) =\frac 18\ln (3\cdot 5^3)\ .\qquad\text{ So:} \\ \lim x_n &= \lim \exp y_n \\ &=\exp\lim y_n\\ &=(3\cdot 5^3)^{1/8}=375^{1/8}\approx 2.09775086317869667712859368405659872\dots\ . \end{aligned} $$

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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.

Answer 2

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.

Answer 3

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}