I have been thinking about this thing for quite a while now. I have tried many ways but couldn't really figure it out.

I am pretty sure that this kind of expressions don't really have a closed form, so I am asking for other representation for example maybe in terms of an infinite sum or product or something.

Creating A New Notation

Just like there is a notation for a sum $\textstyle\displaystyle{\sum_{n=a}^{b}s_n=s_a+\cdots+s_b}$ and also for a product $\textstyle\displaystyle{\prod_{n=a}^{b}s_n=s_a\cdots s_b}$, I was quite surprised that there wasn't any notation for exponentiation.

I would agree that there wouldn't be any use for this notation but still, why would no mathematician ever would create such a notation just for the sake of curiosity. That is why I would request readers to give me any references if there are any. I haven't found any, so I am creating my own. Let $$\boxed{\textstyle\displaystyle{{\huge\varepsilon\normalsize}_{k=a}^{b}s_k=s_a^{\unicode{x22F0}^{s_b}}}}$$ where $b>a$. If $a>b$ then $\textstyle\displaystyle{{\huge\varepsilon\normalsize}_{k=a}^{b}s_k=1}$ and $\textstyle\displaystyle{{\huge\varepsilon\normalsize}_{k=a}^{b}x=^{b-a+1}x}$. Obviously $a,b\in\mathbb{Z}$.

Unlike product and sum, exponentiation isn't commutative so we have to be careful when using the notation. Maybe we can modify it a little bit to include the ordering.

By the way we are defining $\textstyle\displaystyle{{\huge\varepsilon\normalsize}_{k=a}^{\infty}s_k:= \lim_{n\rightarrow\infty}\left({\huge\varepsilon\normalsize}_{k=a}^{n}s_k\right)}.$

Some Natural Questions

When written out in the form of this notation, some natural curious questions arrive or at least some arrived in my mind, for example $\textstyle\displaystyle{{\huge\varepsilon\normalsize}_{k=2}^{\infty}\frac{1}{k^s}}$ and $\textstyle\displaystyle{{\huge\varepsilon\normalsize}_{k=2}^{\infty}\frac{1}{n}}$.

My Curiosity

My initial curiosity was $H=\textstyle\displaystyle{{\huge\varepsilon\normalsize}_{k=2}^{\infty}\frac{1}{k}=\left(\frac{1}{2}\right)^{\left(\frac{1}{3}\right)^{\left(\frac{1}{4}\right)^{\unicode{x22F0}}}}}$.

As pointed out by Tavish in the comments, this can be written as a recurrence relation given by $$\textstyle\displaystyle{a_{n+1}=-\frac{\ln(a_n)}{\ln(n)}}$$ where $\textstyle\displaystyle{a_n=\left(\frac{1}{n}\right)^{\left(\frac{1}{n+1}\right)^{\unicode{x22F0}}}}$. Solving this will help us derive $H$.

But as pointed out in the comments and in this question (Note that this question focuses on the convergence of $H$, while my question focuses on something different), $H$ doesn't really make much sense by the definition of infinite power tower above because it seems that $$\textstyle\displaystyle{\lim_{n\rightarrow\infty}\left({\huge\varepsilon\normalsize}_{k=2}^{2n}\frac{1}{k}\right)\neq\lim_{n\rightarrow\infty}\left({\huge\varepsilon\normalsize}_{k=2}^{2n+1}\frac{1}{k}\right)}.$$ In particular, we have


Let $E_n=\textstyle\displaystyle{{\huge\varepsilon\normalsize}_{k=2}^{2n}\frac{1}{k}}$ and $O_n=\textstyle\displaystyle{{\huge\varepsilon\normalsize}_{k=2}^{2n+1}\frac{1}{k}}$. I tried constructing a recurrence relation for $E_n$ and $O_n$, but couldn't, except \begin{align}E_{n+1}&=\left(\log_{\frac{1}{2n-1}}(\cdots\log_{\frac{1}{2}}(E_n))\right)^{\left(\frac{1}{2n+1}\right)^{\left(\frac{1}{2n+2}\right)}}\\O_{n+1}&=\left(\log_{\frac{1}{2n}}(\cdots\log_{\frac{1}{2}}(O_n))\right)^{\left(\frac{1}{2n+2}\right)^{\left(\frac{1}{2n+3}\right)}}\end{align} which is not really workable. If there is a way to simplify it then please do tell me or write a partial answer about it, because it would help us a lot in finding the values of $H_E$ and $H_O$.

My Question

I am pretty sure that there really isn't closed form of $H_E$ and $H_O$.

So I am asking for a different representation for those constants, maybe as a sum or an integral possibly?

  • 1
    $\begingroup$ I do not think that showing the convergence is easy, but I would be very surprised if this power tower could be expressed in a closed form. $\endgroup$
    – Peter
    Dec 27, 2020 at 12:45
  • 1
    $\begingroup$ The sequence seems to oscillate between $0.65836559926633\cdots$ and $0.6903471261149643\cdots$ for large $n$, so probably the limit does not exist. $\endgroup$
    – Peter
    Dec 27, 2020 at 12:49
  • 1
    $\begingroup$ around $n=300$ with pari/gp , but for $n=1000$ and $n=1001$ the values are almost identical , so we seem to have two partial sequences converging. $\endgroup$
    – Peter
    Dec 27, 2020 at 12:52
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    $\begingroup$ If $$a_n = \left(\frac 1n\right)^ { \left( \frac{1}{n+1} \right)^{\dots}}$$, then $$a_{n+1} = \frac{\ln a_n}{\ln(1/n)} $$ and the goal is $a_2$. Can this recurrence be solved ? $\endgroup$
    – Vishu
    Dec 27, 2020 at 13:14
  • 2
    $\begingroup$ See even and odd OEIS entries. $\endgroup$ Oct 5, 2021 at 17:04

3 Answers 3


It is possible to write the expression as a series. Help on a literature search from a Discord maths server led to the following paper

Bender CM, Vinson JP. Summation of power series by continued exponentials. J. Math. Phys. 37, 4103 (1996).

where the function $a_0\exp(a_1z\exp(a_2z\exp(\cdots)))$ was discussed. A Taylor series expansion is given in equation (1.3) as $$a_0+\sum_{n\ge1}\sum_{j_1+\cdots+j_k=n}\left(\prod_{i=1}^k\frac{a_0(a_ij_{i-1})^{j_i}}{j_i!}\right)z^n$$ where $j_0=1$ and $j_i>0$ for each $i\ge1$ (a multinomial sum of sorts). Evidently, given a particular value of $z$, this function converges if the sequence of exponential convergents $a_0,a_0\exp(a_1z),\cdots$ converges. However, as shown in Convergence of $a_n=(1/2)^{(1/3)^{...^{(1/n)}}}$, our sequence converges to two distinct values subsequenced by parity, so the results in the paper do not immediately apply.

Considering the even case first (the sequence $E_n$ in your notation), we can get round this by taking the function $a_0\exp(a_1\exp(a_2z\exp(a_3\exp(a_4z\exp(\cdots)))))$ instead, where the indeterminate $z$ occurs only at even indices of $a_m$. The coefficients are such that $a_0=1$ and $a_m=-\log(m+1)$ for each $m\ge1$ and eventually we will take $z=1$ to evaluate the expression.

Writing $f_k(z)=\exp(a_{2k+1}\exp(a_{2k+2}zf_{k+1}(z)))$ for each $k\ge0$, we obtain \begin{align}\small f_0(z)&\small=e^{a_1}\left(1+a_1\left(a_2zf_1(z)+\frac{a_2^2}2z^2f_1(z)^2+\cdots\right)+\frac{a_1^2}2\left(a_2zf_1(z)+\frac{a_2^2}2z^2f_1(z)^2+\cdots\right)^2+\cdots\right)\\\small f_1(z)&\small=e^{a_3}\left(1+a_3\left(a_4zf_2(z)+\frac{a_4^2}2z^2f_2(z)^2+\cdots\right)+\frac{a_3^2}2\left(a_4zf_2(z)+\frac{a_4^2}2z^2f_2(z)^2+\cdots\right)^2+\cdots\right)\end{align} and so on. The first few coefficients of $z^n$ in $f_0(z)$ are as follows \begin{align}[z^0]&=e^{a_1}\\ [z^1]&=e^{a_1}a_1a_2e^{a_3}\\ [z^2]&=e^{a_1}a_1a_2e^{a_3}a_3a_4e^{a_5}+e^{a_1}a_1\frac{a_2^2}2e^{2a_3}+e^{a_1}\frac{a_1^2}2a_2^2e^{2a_3}\\ [z^3]&=\tiny e^{a_1}a_1a_2e^{a_3}\left(a_3\frac{a_4^2}2e^{2a_5}+\frac{a_3^2}2a_4^2e^{2a_5}\right)+e^{a_1}a_1\frac{a_2^2}2\cdot2e^{a_3}a_3a_4e^{a_5}+e^{a_1}a_1\frac{a_2^3}6e^{3a_3}+e^{a_1}\frac{a_1^2}2a_2^3e^{3a_3}+\frac{a_1^3}6a_2^3e^{3a_3}.\end{align} We note that the coefficients are much less elegant that those in the function considered in the paper, because dropping the indeterminate $z$ results in irregular contributions when extracting each term. I don't have time at this moment to attempt a closed form expression, but it seems likely that there is one (in the form of a double multinomial sum). And from which we immediately have $H_E=\sum\limits_{i\ge0}[z^i]$.

We can proceed with the odd case in a very similar way by taking the function $\exp(a_1z\exp(a_2\exp(a_3z\exp(\cdots))))$ and the sum of its coefficients will give us $H_O$.

  • $\begingroup$ I'm slightly confused : the original sequence $\frac 12^{\frac 13^ \ldots}$ has two distinct limit points, so how can it equal one expression? It must be the case that the partial sums of the infinite sum found here, have two different points. I'll consult the paper. $\endgroup$ Oct 9, 2021 at 17:59
  • $\begingroup$ @TheSimpliFire That's what I thought as well, thanks. The paper makes no real reveal about it as I see it, but I think this is what is supposed to occur. This can be used to prove that $H_E \neq H_O$ if you can prove that every term in the summation (in absolute value) stays above a certain $C>0$. I haven't seen anywhere in the literature, a proof that $H_E \neq H_O$, but the MSE proof above seems to be one. $\endgroup$ Oct 9, 2021 at 18:44
  • $\begingroup$ @TeresaLisbon See revised. $\endgroup$
    – TheSimpliFire
    Oct 10, 2021 at 10:53
  • $\begingroup$ @TheSimpliFire This is exactly what I expected, thanks. $\endgroup$ Oct 10, 2021 at 11:19
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    $\begingroup$ @TheSimpliFire. This perfectly answers my question,I have selected your answer, but the bounty will be given only when the bounty period ends. $\endgroup$ Oct 10, 2021 at 11:30


Using your notation, define $\overset{n}{\underset{k=2}{\huge \varepsilon}} \, \frac{1}{k} = A_n$ if $n$ is odd and $B_n$ if $n$ is even. We can subtract the limits from each "branch" and flip the lower one to be able to perform one regression on the logarithm of the sequence:

For $n > 10$ the sequence is approximated pretty well by

$$ f(n) = \exp (-0.0034n^2 - 0.117n - 2.912) \qquad \to \begin{matrix} A_n = f(n) + A_\infty \\ B_n = B_\infty - f(n) \end{matrix} $$

By the way, the recursion is pretty useless because it starts at $n = \infty$ and $a_2 = - \frac{\ln a_1}{\ln 1}$ which is not determined.

  • $\begingroup$ Which software did you use for those beautiful plots? $\endgroup$ Oct 7, 2021 at 20:03
  • $\begingroup$ @Christian Is there an alternate form of the sequence? $\endgroup$ Oct 7, 2021 at 20:25
  • $\begingroup$ @principal-ideal-domain I used Python with Matplotlib. $\endgroup$
    – Christian
    Oct 8, 2021 at 0:32

I will use my same answer as in Christian’s post. Better a form that none. This will be an attempt at a semi closed form by using logarithm properties. The fraction MathJax will not be used for better viewing let this number, odd or even limit, be called C for constant. Note that these represent both cases of $C=\text H_{e,o}$: $$C=(1/2)^{{1/3}^{{...}^{1/n}}}=\exp\biggr(\ln\biggr((1/2)^{{1/3}^{{...}^{1/n}}}\biggr)\biggr)= \exp\bigr({{1/3}^{{...}^{1/n}}} \cdot -\ln(2)\bigr)= \exp\bigr(\exp\bigr({1/4}^{{...}^{1/n}} \cdot-\ln(3)\bigr) \cdot-\ln(2)\bigr)=\exp(\exp(…(-\ln(n)) \cdot-\ln(n-1))… \cdot-\ln(3)) \cdot-\ln(2))$$

This means our final answer is: $$C=\lim_{n\to \infty} \exp(\exp(…(-\ln(n)) \cdot-\ln(n-1))… \cdot-\ln(3)) \cdot-\ln(2)) = \lim_{n\to \infty} e^{e^{e^{{.^{.^.}}^{(-\ln(n)) \cdot-\ln(n-1)}…\cdot-\ln(3)} \cdot-\ln(2)}}$$

Here is an attempt at a differential equation using a similar process. I will differentiate the general version which may combine both cases:

$$y(x)=2^{{-3}^{{-4}^{.^{.^{.^{{(x-1)}^x}}}}}}= 2^{{-3}^{{-4}^ {{.^{.^.}}^{g(x)}}}}$$

Let’s differentiate to try and find a differential equation:

$$y’=y’(x)=\frac d{dx} \exp \ln\left( 2^{{-3}^{{-4}^ {{.^{.^.}}^{g(x)}}}}\right) =\frac d{dx}\exp\left((-3)^{{-4}^{{-5}^ {{.^{.^.}}^{g(x)}}}}\ln(2)\right)=2^{-3^{{-4}^{{-5}^ {{.^{.^.}}^{g(x)}}}}}\ln(2)\frac d{dx} -\exp\left( {{-4}^{{-5}^ {{.^{.^.}}^{g(x)}}}}\ln(3)\right)= -2^{-3^{{-4}^{{-5}^ {{.^{.^.}}^{g(x)}}}}} 3^{-4^{{-5}^{{-6}^ {{.^{.^.}}^{g(x)}}}}}\ln(2)\ln(3)\frac d{dx} 4^{{-5}^{{-6}^ {{.^{.^.}}^{g(x)}}}}=-y\log_2(y)\ln(2)\ln(3)\frac d{dx} \log_2(\log_3(y))=-\ln(2)y\frac d{dx}\log_2(y)= \ln(2)y\frac{y’(x)}{\ln(2)y}=y= … $$

However, this produces a true expression. Trying to differentiate @Tavish’s $a_n=\left(\frac1n\right)^{\left(\frac1{n+1}\right)^…}$ produces an increasingly complicated derivative, so no functional equation not differential equation can be made so far to try and find a value of $\text H_{e,o}$.

Let me think of something else. This also looks like another recurrence relation. Please correct me and give me feedback!

  • $\begingroup$ This is just a reformulation of the problem. $\endgroup$
    – Christian
    Oct 8, 2021 at 8:43

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