$\sqrt[3]{8\div{\sqrt[3]{8\div{\sqrt[3]{8\div{\sqrt[3]{8\div ...} }} }} }} $=?

Question

Suppose $$a=\sqrt[3]{8\div{\sqrt[3]{8\div{\sqrt[3]{8\div{\sqrt[3]{8\div ...} }} }} }} $$ $\bf{Question}:$Is it possible to find the value of $a$

Thanks in advance for any hint,idea or solution.

$\bf{remark}:$ I changed the first question , But I got stuck on this .

Answer 1

Consider the sequence,

$$x_{n+1}=\sqrt[3]{\frac{8}{x_n}}=2(x_n)^{-\frac{1}{3}}$$

With $x_1=1$. Our value of interest is $\lim_{n \to \infty} x_n$.

Such a sequence follows,

$$\ln x_{n+1}=\ln 2-\frac{1}{3} \ln x_n$$

Hence letting $\ln x_n=a_n$ we have the linear recurrence,

$$a_{n+1}+\frac{1}{3}a_{n}=\ln 2$$

$$(a_{n+1}-\frac{3}{4}\ln 2)+\frac{1}{3}(a_n-\frac{3}{4}\ln 2)=0$$

One may show $a_n-\frac{3}{4} \ln 2 \to 0$ in much the same way $(-\frac{1}{3})^n \to 0$. For instance, by first finding a closed form by letting $b_n=a_n-\frac{3}{4}\ln 2$. Hence $a_n \to \frac{3}{4}\ln 2$, thus showing that $x_n \to 2^{3/4}$.

Answer 2

This is equivalent to summing an infinite GP. Converting the nested radicals into fractional exponents we can see that the exponent of $8$ is $$\frac{1}{3}-\frac{1}{3^{2}}+\frac{1}{3^{3}}-\dots=\frac{1/3}{1+(1/3)}=\frac{1}{4}$$ and therefore the given expression equals $8^{1/4}=2^{3/4}$.