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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

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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.

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    $\begingroup$ @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. is a poor mathematician Aug 31 '11 at 20:12
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    $\begingroup$ Technically, you need to prove convergence before you can do that. $\endgroup$ – GEdgar Aug 31 '11 at 20:36
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    $\begingroup$ AHA! precalculus mean we don't need no stinkin' proofs ?? $\endgroup$ – GEdgar Aug 31 '11 at 21:07
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    $\begingroup$ $Y^2=1+Y$ has $Y=\infty$ as solution, as well as the one you found... $\endgroup$ – GEdgar Aug 31 '11 at 21:09
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    $\begingroup$ Related: math.stackexchange.com/questions/11945/… $\endgroup$ – Aryabhata Sep 1 '11 at 15:16

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