\[\sqrt[n]{x \sqrt[n]{ y \sqrt[n]{x \sqrt[n]{y \sqrt[n]{x \sqrt[n]{y \sqrt[n]{x \cdots }}}}}}}\]

I noticed something looking at the convergence of geometric series as they relate the questions of the form shown above.

My brain is more empirical and inductive, so I am not as gifted with proofs as the rest of you, nor do I find it easy to tell whether or not something I've found is completely obvious. But what I have done is calculate the convergence of both variables separately. Just using the series test, you can prove convergence, and use any method to determine the actual convergence of a series. I found the series to be in a form: \(\frac{1}{x^{n+(n-1)}}\), for example. I simply found a pattern that it seems to stick to, and I believe that just proving that all the series converge is proof enough. Anyway here is the result:

\[\sqrt[n]{x \sqrt[n]{ y \sqrt[n]{x \sqrt[n]{y \sqrt[n]{x \sqrt[n]{y \sqrt[n]{x \cdots}}}}}}}= x^{\frac{n}{q}}\cdot y^{\frac{1}{q}}\] Where \(\displaystyle q=\sum_{i=2}^{n} n+(n-1)\).

For example: \[\sqrt{3\sqrt{2\sqrt{3\sqrt{2 \cdots}}}}=3^{\frac{2}{3}}\cdot 2^{\frac{1}{3}}\]

Like I said, I don't know if this is obvious. Somebody probably has already discovered this before, but maybe someone will find it useful.

## Comments

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TopNewestGood note !!

More or less the same, we can also say that,

\( \begin{equation} \begin{split} \sqrt[n]{x \sqrt[n]{ y \sqrt[n]{x \sqrt[n]{y \sqrt[n]{x \sqrt[n]{y \sqrt[n] {x \cdots}}}}}}} & =x^ay^b \\ \Rightarrow \begin{cases} a &=\frac{\frac{1}{n}}{1-\frac{1}{n^2}}=\frac{n}{n^2-1} \\ b & = \frac{\frac{1}{n^2}}{1-\frac{1}{n^2}}=\frac{1}{n^2-1} \end{cases} \end{split} \end{equation}\) – Akshat Sharda · 7 months, 3 weeks ago

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@Akshat Sharda ! It is very elegant in terms of procedure. It probably makes the calculation a bit easier. I am curious to know your reasoning behind it, if you don't mind. :) – Drex Beckman · 7 months, 3 weeks ago

Brilliant solution ,Log in to reply