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Let $k \in \mathbb{R}$; for what $k$ does

$$ \lim_{n \to \infty} k \uparrow\uparrow n $$

converge?

Note: The double up-arrow refers to Knuth’s up-arrow notation, so $k \uparrow\uparrow n$ means $n$ levels of $k^{k^{\dots}}$.

Franklin Yu
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    See here for the precise result. – Gary May 12 '22 at 04:33
  • The explanation of Knuth's notation is short enough to be included in your text. Avoid asking the reader to go to an entire article. – Kurt G. May 12 '22 at 04:33
  • To be precise: is $k\uparrow\uparrow 2$ meant to be $(k^k)^k$ or is it $k^{(k^k)}$ ? – Kurt G. May 12 '22 at 05:59
  • @KurtG. Power towers are always calculated from above (from right), this is a convention. The reason is that this way we usually get the largest possible number. – Peter May 12 '22 at 08:11
  • This has already been asked here (see the above comment). – Peter May 12 '22 at 08:12
  • @Peter . I knew the answer. My point was that there is a very inexpensive an reader-friendly way to clarify that notation every time it is introduced. But that view might be opinion based. – Kurt G. May 12 '22 at 08:38
  • Thanks Gary! Closed this question. – Franklin Yu May 12 '22 at 16:50

1 Answers1

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This answer only covers the case of positive $k$. For $k \in (0, 1]$ the limit trivially converges. The case $k > 1$ is covered below.

finding upper bound

Assuming the limit converges to a real number $x$, we have

\begin{align} k^x &= x \\ x \ln k &= \ln x \\ \ln k &= \frac{\ln x}{x} =: f(x) \end{align}

The function $f$ looks like this. WolframAlpha tells us that

$$ \max_{x > 0} f(x) = f\left( \frac{1}{e} \right) = \frac{1}{e} $$

Therefore the upper bound of $k$ is

$$ k_0 = e^{1/e} \approx 1.445 $$

proof

Now we need to prove our assumption above, i.e. the limit does converge within the entire interval $(1, k_0]$. First pick any $k$ in the interval. Let $x_0$ be a solution to $k^x = x$ with the given $k$; from the plot we can see that $k > 1 \implies x_0 > 1$. Then, construct a sequence $\{a_n\}$ such that $a_0 = 1$, and $a_{i+1} = k^{a_i}$ (in other words, $a_n = k \uparrow\uparrow n$) for all natural number $i$. The first few terms look like

$$ a_1 = k, \quad a_2 = k^k, \quad a_3 = k^{k^k}, \quad \ldots $$

Since $a_0 = 1 < x_0$, and for any $i \in \mathbb{N}$

$$ a_i < x_0 \quad \implies \quad a_{i+1} = k^{a_i} < k^{x_0} = x_0 $$

By induction we know that the entire sequence $\{a_n\}$ is strictly smaller than $x_0$. In addition, we know that $a_1 = k > 1 = a_0$, and for any positive $i$ we have

$$ a_i > a_{i-1} \quad \implies \quad a_{i+1} = k^{a_i} > k^{a_{i-1}} = a_i $$

By induction we know that the sequence $\{a_n\}$ is strictly increasing. In other words, the sequence is bounded in interval $[1, x_0)$. Because the metric space for $\mathbb{R}$ (with the canonical metric and order operator) is compact, monotonic and bounded sequence converges. Q.E.D.

Franklin Yu
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