How would one go about solving in this equation? I am more used to solving equations in this form:


Which you solve in this form:


But how would you solve that equation at the top of the page though? If you were curious, and I know that SE likes what I have tried, so I will show steps that I have attempted.



But from here I am not sure what to do from here. Can someone please help me evaluate this equation for $a$ in therms of $n$? Can the proof of this also be somewhat rigorous please?

  • $\begingroup$ Does $a^{(a+1)^{\dots}}$ converge if $|a|>1$? $\endgroup$ – TomGrubb Jul 18 '15 at 23:34
  • $\begingroup$ @bburGsamohT Certainly not. Consider the sequence $|a_0|>1$ and $a_n=(a_{n-1})^n$. Since $\forall \{b,c\}$ such that $|b|>1$ and $c>1$, necessarily $|b^c|>b$, we can easily induct to find that $\lim_{n \to \infty} a_n$ is at the very least infinite. $\endgroup$ – Archaick Jul 18 '15 at 23:46
  • $\begingroup$ Where is $a$ chosen from? Is it a positive real number? $\endgroup$ – A.P. Jul 18 '15 at 23:56
  • $\begingroup$ @Archaick right, thanks, that's what I was kinda thinking. So it seems like there will be three cases: $|a|<1$, in which case you get $0$, $|a|=1$, in which case you get $1$, and $|a|>1$, in which case it diverges. So it doesn't seem like there will be much to this. $\endgroup$ – TomGrubb Jul 19 '15 at 0:12
  • $\begingroup$ @Archaick But if $a\in (-1,0]$, then $a+1$ is positive, and thus $|a^{a+1}|=|a|^{a+1}<|a|$, right? Every successive power should decrease the magnitude Edit: not quite right, you have to go out to $a+2>1$, and then the magnitude should start decreasing. It will probably be that $|a^{a+1}|>a$, but we will still have $|a^{a+1}|<1$ which is all that matters $\endgroup$ – TomGrubb Jul 19 '15 at 0:21

We need additional resources to handle "general" sequences of infinite exponentials. In particular:

$\mathbf{Definition:}$ Suppose $A_k=\{a_1,a_2,\ldots,a_k\}$, $k\in\mathbb{N}$, with $a_k>0$ and $n\le |A_k|=k$. A general infinite exponential is:

$$e_n(A_k) = \begin{cases} a_k, & \text{if $n=1$} \\ a_{k-n+1}^{e_{n-1}(A_k)}, & \text{if $n>1$} \end{cases}$$

Now, if one sets $b_n=e_n(A_n)$, then the sequence $b_n$, $n\in\mathbb{N}$ expresses the sequence of ascending exponentials:


Now you need the following theorem:

$\mathbf{Theorem\,\,(Barrow):}$ The sequence $b_n$, $n\in\mathbb{N}$ converges iff:

  • $a_n$ converges
  • $\exists n_0:\forall n>n_0:b_n\in [e^{-e},e^{1/e}]$

From the above theorem it is clear that your example is unsolvable, since your sequence is $a_n=a_{n-1}+1$, with $a_1=a$ and this sequence diverges, therefore it fails bullet one.

Suppose instead that you are called to solve, using an $a_n$ which satisfies the theorem's assumptions:

$$y=a_1^{a_2^{a_3^{\cdots}}}\Rightarrow$$ $$\ln(y)=\ln(a_1)\cdot a_2^{a_3^{a_4^{\cdots}}}\Rightarrow$$ $$\frac{\ln(y)}{\ln(a_1)}=a_2^{a_3^{a_4^{\cdots}}}$$

Now, $a_n$ converges, so if we fix $\epsilon>0$ we are guaranteed a $k>0$ such that for all $n>k$, we have $a_n \sim a$, where $a=\lim\limits_{n\to\infty}a_n$.

We can therefore continue the iteration with logarithms as above, all the way to $k$:


which is equivalent within $\epsilon$, to:

$$Y_{\epsilon}=\frac{(\cdots)}{\ln(a_k)}\sim a^{a^{a^{\cdots}}}$$

the latter being now solvable using the trick you mention in your question (or any other valid method) either for $a$ or for $Y_{\epsilon}$. In particular:

$$a\sim Y_\epsilon^{\frac{1}{Y_{\epsilon}}}\Leftrightarrow Y_{\epsilon}=\frac{W(-\ln(a))}{-\ln(a)}$$

where $W$ is the Lambert function.


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