If $\lim_{n\to \infty} a_n = 4$, can we say $\lim_{n\to \infty} a_{n-2} = 4$?

Question

I was doing this exponential tower equation: $$2^{x^{2^{x^{2^{...}}}}} = 4$$ $$\text {(each new exponent is the power of the last exponent)}$$

The popular method is to break the tower at the first exponent and evaluate that what's left in the exponential tower is still the LHS if the tower length goes into infinity: $$2^{x^{2^{x^{2^{...}}}}} = 2^{x^{4}} = 4$$

Consider sequence $a_n$ with $a_1 = 2^x$, $a_2 = 2^{x^{2}}$, $a_3 = 2^{x^{2^{x}}}$, and so on, then:

$$LHS = \lim_{n\to ∞} a_n = 4$$

The method above seems to be reasoning that: $$\text{if}\;\;\;\;\;\;\;\;\;\;\lim_{n\to ∞} a_n = 4$$

$$\text{then}\;\;\;\;\;\;\;\;\;\lim_{n\to ∞} a_{n-2} = 4$$

(1) Is this something that we need to prove? I know this is intuitive enough, but is there a way to rigorously demonstrate this?

(2) I was assuming that this is the reasoning behind the method above. If you think otherwise, please advise!

(3) Unrelated, but is there a formula to define the sequence above with $a_n, a_{n+1}, a_{n+2}$? (I assume we cannot write a general case for $a_n$ alone.)

Thank you!

Answer 1

Concerning (1), this is indeed true and very intuitive. You would probably only prove it in your first years studying mathematics (after that, such a basic fact does not require a proof).

Given a sequence $(u_n)_{n\in\mathbb{N}} \in \mathbb{R}^\mathbb{N}$ and $\ell \in \mathbb{R}$, the definition of $\lim_{n\to+\infty} u_n = \ell$ is that: for every $\varepsilon> 0$, there exists $N \in \mathbb{N}$, such that, for every $n \geq N$, $|u_n-\ell|\leq \varepsilon$.

If you are familiar with quantifiers, it should be fairly easy for you to prove (1) by applying this definition to your assumption on $u_n = a_n$, and show that it entails the same limit for $u_n = a_{n-2}$.

Answer 2

So, in general, your question seems to be

Suppose $\displaystyle \lim_{n \to \infty} a_n = L$. Is it true that $\displaystyle\lim_{n \to \infty} a_{n-s} = L$ for a fixed $s$?

This is, indeed, true.

The simplest angle of argument would be the fact that, if $a_n \to L$, then any subsequence $a_{n_k} \to L$. The case of a "shifted" sequence $a_{n-s}$ (truncating off the first few terms so that this new sequence is well-defined) is a special case of this: $n_k := k-s$.

One could also argue this formally. We recall that (for $L$ finite, I leave the infinite cases to you)

$$a_n \to L \iff (\forall \varepsilon > 0)(\exists N \in \mathbb{N})(\forall n \ge N) \Big(|a_n-L| < \varepsilon \Big)$$

(If you're unfamiliar: "$\forall$" means "for all" and "$\exists$" means "there exists".)

Then we can see that $a_{n-s} \to L$ as well, by using the integer $N+s$.

---

For your second question: I don't really think the recursion is the intention.

It is a common theme with problems of this type to have a "self-similar" expression, e.g.

$$y = x^{x^{x^{x^\cdots}}} \implies y = x^y$$

or

$$\alpha = \frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\ddots}}}} \implies \alpha = \frac{1}{1+\alpha}$$

The end result is reducing the expression to one purely algebraic, without an implied limiting process (albeit with the assumption that the limiting process indeed has a limit!).

When we hit $2^{x^4}=4$, we can solve for $x$ by seeing

$$x^4 = \log_2(4) \implies x = \sqrt[4]{\log_2(4)}$$

(if we strictly want a real positive root) -- no appeal to recursion necessary here. (Though we may need to appeal to recursion to prove that this limit exists in the first place -- but for nicely-behaved such problems where the limit exists, we can do this.)

---

To answer your final question, recursion would probably be the best way, but one need be careful of order of exponentiation and exponentiation properties. I think you would be stuck with letting $a_{n+1} = 2^{a_{n}}$ or $a_{n+1} = x^{a_n}$ depending on the sign of $n$; the details I would leave to you.

As far as a meaningful explicit formula (that doesn't just resort to obfuscation of a much simpler "by cases, depending on the sign of $n$" definition), no clue.

Answer 3

Is this something that we need to prove?

Of course. Is there anything in maths that we don't need to prove? Maybe only those things that you already proved. :)

More generally, assume $x_n\to x$ in some (topological) space $X$. Let $b_n$ be a sequence of naturals such that $b_n\to\infty$. Then $x_{b_n}\to x$.

Proof. Let $U$ be an open neighborhood of $x$. Then there is $N$ such that $x_n\in U$ for all $n>N$. Furthermore there is $M$ such that $b_m>N$ for all $m>M$. And that means $x_{b_m}\in U$ for $m>M$. The arbitrary choice of $U$ completes the proof. $\Box$