How would one go about computing the value of $X$, where

$X=5^0+ \sqrt{5^1+\sqrt{5^2+\sqrt{5^4+\sqrt{5^8+\sqrt{5^{16}+\sqrt{5^{32}+\dots}}}}}}$

I have tried the standard way of squaring then trying some trick, but nothing is working. I have also looked at some nested radical previous results, but none seem to be of the variety of this problem. Can anyone come up with the answer? Thank


The trick is to pull out a $\sqrt{5}$ factor from the second term:
$$ \frac{\sqrt{5^1+ \sqrt{5^2 + \sqrt{5^4 + \sqrt{5^8 + \cdots}}}}}{\sqrt{5}} = \sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}}, $$ which I call $Y$ for convenience. To see why this is true, observe that $$ \begin{align*} \frac{\sqrt{5^1+x}}{\sqrt{5}} = \sqrt{1+\frac{x}{5^1}} \\ \frac{\sqrt{5^2+x}}{5^1} = \sqrt{1+\frac{x}{5^2}} \\ \frac{\sqrt{5^4+x}}{5^2} = \sqrt{1+\frac{x}{5^4}} \end{align*} $$ and so on. Applying these repeatedly inside the nested radicals gives the claim. (The wording of this explanation has been inspired mostly by a comment of @J.M. below.)

Now, it only remains to compute $Y$ and $X$. By squaring, we get $$ Y^2 = 1 + Y \ \ \ \ \implies Y = \frac{\sqrt{5}+1}{2}, $$ discarding the negative root. Plugging this value in the definition of $X$, we get: $$ X = 1 + \sqrt{5} Y = 1 + \frac{5+\sqrt{5}}{2} = \frac{7+\sqrt{5}}{2} . $$

Note on the convergence issues. As @GEdgar points out, to complete the proof, I also need to demonstrate that both sides of the first equation do converge to some finite limits. For our expressions, convergence follows from @Bill Dubuque's answer to my question on the defining the convergence of such an expression. I believe that with some work, one can also give a direct proof by showing that this sequence is bounded from above (which I hope will also end up showing the theorem Bill quotes), but I will not pursue this further. Added: See @Aryabhata's answer to a related question for a hands-on proof.

  • 4
    $\begingroup$ @Fool: you mean the nested radical? Note that $\sqrt{5+x}/\sqrt 5=\sqrt{1+x/5}$; just keep applying that as you go deeper. $\endgroup$ – J. M. isn't a mathematician Aug 31 '11 at 20:12
  • 3
    $\begingroup$ Technically, you need to prove convergence before you can do that. $\endgroup$ – GEdgar Aug 31 '11 at 20:36
  • 2
    $\begingroup$ AHA! precalculus mean we don't need no stinkin' proofs ?? $\endgroup$ – GEdgar Aug 31 '11 at 21:07
  • 2
    $\begingroup$ $Y^2=1+Y$ has $Y=\infty$ as solution, as well as the one you found... $\endgroup$ – GEdgar Aug 31 '11 at 21:09
  • 1
    $\begingroup$ Related: math.stackexchange.com/questions/11945/… $\endgroup$ – Aryabhata Sep 1 '11 at 15:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.