Tetration limit

Question

For each $n$, define $f_n:\mathbb R^+\rightarrow \mathbb R^+$ by $f_n(x) = \underbrace{x^{x^{x^{...^{x^x}}}}}_n$

Is it true that $\lim\limits_{n \to \infty} f_n(\frac{n+1}{n}) = 1$ ?

A few computations using Wolfram Alpha seem to suggest so, but I am unsure of how to prove it.

One way to think about why it might be true intuitively is to compare it to the limit definition of $e$, but even this intuition is handwavy (although it seems like it is true based on computations).

Furthermore, I'm unsure if the Binomial expansion actually helps us here.

I'm interested in hearing your thoughts.

Answer 1

The answer is yes. In the Wikipedia article there is a graph, which gives $$\lim_{n\rightarrow \infty} f_n(x)$$

For $e^{-e}<x<e^{1/e}$, this limit exists and is between $e^{-1}$ and $e$. As $x\rightarrow 1$, the limit approaches 1, just as conjectured by OP.

Answer 2

Tetration to infinite height converges in the range $\left[e^{-e} , e^{\frac1e}\right]$ and nowhere else. See Wikipedia for discussion of this, but it's easy enough to see that this restriction must be true if the definition $\displaystyle{e}=\lim_{n\rightarrow\infty}{\left(\frac{n+1}{n}\right)}^n$ is to hold.

Answer 3

The answers above are all correct.

I believe there is an intuitive way to see $why$ Tetration to infinite height converges for $ x \in [e^{-e},e^{1/e}]$ (although admittedly it is extremely handwavy). I provide and intuitive argument for the above within the range $x<e^{1/e}$. The same argument can probably be applied for $x>e^{-e}$.

Bold Statement: Given $x$ less than but close to $e^{1/e}$, a solution (for $y$) to the equation

$ x^y = y$

exists and is less than $e$.

Take, for example, $1.4 < e^{\frac1e}$.

http://www.wolframalpha.com/input/?i=1.4%5Ex+%3D+x

The solution to the equation $1.4^y = y$ we are interested in is 1.88666... .

Trying this for values closer and closer to (but less than) $e^{1/e}$ suggests the "Bold Statement" above, although I haven't really thought about a proof.

We can summarise the Bold Statement by saying that for any given $x<e^{1/e}$ (and with x close to $e^{1/e}$), $f(y) = x^y$ has a "fixed point".

Applying the Contraction Mapping Theorem to $f(y)$ will (hopefully) yield the result which solves the OP.

Answer 4

Your argument $ {n+1 \over n} $ can be rwritten as $ 1+ \frac 1n $ and then one could look at $\lim_{x\to 0} 1+x $ where $x = \frac 1n$. .

Then observe the sequence of Taylor-series of

$f(x)=1+x$,

$f_2(x)=(1+x)^{1+x} = 1 + x + x^2 + 1/2*x^3 + O(x^4)$,

$f_3(x)=(1+x)^{(1+x)^{1+x}} = 1 + x + x^2 + 3/2*x^3 + O(x^4) $,

and so on. You'll find, that the coefficients at the powers of $x$ stabilize one per each iteration. Now assume $n \to \infty$ which means $x \to 0$ and see that the infinite series reduces to the constant $1$, so this makes it much intuitive that indeed the limit is $1$.

I think it is easy to make it formal by considering the power series of $(1+x)^{1+x}=\exp( \log(1+x) \cdot (1+x + O(x^2) ) )$ and induction.

Answer 5

Let $g(x)$ be the inverse function of $x\to x^{1/x}$, which is strictly increasing on $[1,e]$, hence $g$ exists and is increasing on $[1,e^{1/e}]$. This definition implies that $x^{g(x)} = g(x)$ because $g(x)^{1/g(x)} = x$.

For any $x\in[1,e^{1/e}]$, We can use induction to show $$ f_n(x) \le g(x) $$ for all $n$. For the base case, obviously $f_0(x) = x\le g(x)$. If the inequality holds for $n$, then we have that $$ f_{n+1}(x) = x^{f_n(x)} \le x^{g(x)} = g(x) $$ Therefore, we can conclude by observing $$ 1 \le f_n(\frac{1+n}{n}) \le g(\frac{1+n}{n}) $$ $g$ is continuous, so taking the limit on the RHS, we can see that the squeeze theorem implies the limit of $f_n(\frac{1+n}{n})$ is $1$.