Find $\sqrt[m]{\frac{\sqrt[m]{\frac{\sqrt[m]{\frac{\sqrt[m]{a}}{a}}}{a}}}{\begin{array}{c} a\\\vdots \end{array}}}$

Question

Assuming $m\in \Bbb N\setminus\{0,1\}$ and $a\in \Bbb R_+\setminus\{0\}$, find $$\sqrt[m]{\frac{\sqrt[m]{\frac{\sqrt[m]{\frac{\sqrt[m]{a}}{a}}}{a}}}{\begin{array}{c} a\\\vdots \end{array}}}$$

In a book I'm studying the author gives $$a^{\frac{m}{m+1}}$$ as the answer. But I'm a little uncomfortable with the development presented. In an example, in which $a=128$ and $m=6$, the numeric result given is 64 (by applying the formula).

But, trying to compute a numerical evaluation of the sequence, it appears to converge to something less then 0.38.

Confirmation of the result or a correct answer to the problem will be appreciated. Sorry if this is a duplicate.

Answer 1

As it stands the recurrence is \begin{eqnarray*} A_n= \sqrt[m]{ \frac{A_{n-1}}{a}}. \end{eqnarray*} Now assuming a limit $A$ exist then gives $A^m a=A$, assume $A=a^{\alpha}$ then $ \alpha=1/(1-m)$ and we have the solution \begin{eqnarray*} A= a^{\frac{1}{1-m}} \end{eqnarray*} So in your example $a=128, m=6$ does indeed give $A=0.378929 \cdots$.

Not sure what the recurrence would need to be in order to get the solution suggested by the author.

Answer 2

For $m>1$, $a>0$ and $x_0>0$, the recurrence $x_{n+1}=\frac a{\sqrt [m]{x_n}} $ has limit $\ell $ satisfying $\ell=\frac {\sqrt [m]{\ell}}a $ from which $\ell=a^{\frac m {1+m} }$. Hence for $a=128$ and $m=6$ you get the limit $\ell=64$.