Proof of the inequality $2\uparrow^n 4 < 3\uparrow^n 3 < 2\uparrow^n 5$

Question

I tried to prove the inequality

$$2\uparrow^n 4 < 3\uparrow^n 3 < 2\uparrow^n 5$$

for all natural numbers $n\ge 1$

For $n = 1$, the claim is true because of $16 < 27 < 32.$

The left inequality can be proven with induction

$2 \uparrow^{n+1}4 = 2 \uparrow^n(2\uparrow^{n+1} 3) = 2\uparrow^n (2\uparrow^n4) < 2\uparrow^n(3\uparrow^n 3) < 3\uparrow^n(3\uparrow^n 3) = 3\uparrow^{n+1} 3 $

(Please check this proof)

The right inequality is true for $n = 2$ because of

$$3\uparrow \uparrow 3 \approx 7,6 * 10^{12}$$ $$2\uparrow \uparrow 5 \approx 10^{19728}$$

For $n = 3$ we have

$$3\uparrow \uparrow \uparrow 3 \approx 10 \uparrow \uparrow (7,6*10^{12})$$ $$2\uparrow \uparrow \uparrow 5 \approx 10 \uparrow \uparrow (2 \uparrow \uparrow 65536)$$

So, for $n = 3$ the claim is also true.

It seems clear that for bigger $n$, the inequality also holds, but I am looking for a rigorous proof. Perhaps, the base exchange theorem from r.e.s. can help ?

Answer 1

We can generalize the inequalities to:

$$2 \uparrow^m (n+1) < 3 \uparrow^m n < (2 \uparrow^m (n+2))-2$$

for $m,n \ge 2$. We will prove it by double induction.

For $m=2$, we have by an easy induction

$$2 \uparrow\uparrow (n+2) = 2^{2 \uparrow\uparrow (n+1)} < 2^{3 \uparrow\uparrow n} < 3 ^{3 \uparrow\uparrow n} = 3\uparrow\uparrow(n+1)$$

For the right inequality, we instead prove $4 (3 \uparrow\uparrow n) < 2 \uparrow\uparrow (n+2)$.

Base case: $4(3\uparrow\uparrow 2) = 4*27 = 108 < 2\uparrow\uparrow 4 = 65536$.

Induction:

$$2 \uparrow\uparrow (n+3) = 2^{2 \uparrow\uparrow (n+2)} > 2^{4(3\uparrow\uparrow n)} = 16^{3\uparrow\uparrow n} > 4*3^{3\uparrow\uparrow n} = 4(3\uparrow\uparrow (n+1))$$

Next case is $n=2$. For the left inequality we have:

$$2 \uparrow^m 3 = 2\uparrow^{m-1} 4 < 3\uparrow^{m-1} 3 = 3\uparrow^m 2$$

For the right inequality, we use the fact that $2\uparrow^{m-1}4 > 5$. Then

$$2 \uparrow^m 4 - 2 = 2\uparrow^{m-1}(2\uparrow^{m-1}4)- 2 > 2 \uparrow^{m-1}5 - 2 > 3\uparrow^{m-1} 3 = 3\uparrow^m 2$$

Finally, we have the general case $m,n \ge 3$. For the left inequality,

$$2 \uparrow^m (n+1) = 2\uparrow^{m-1} (2 \uparrow ^m n) < 2 \uparrow^{m-1}(3\uparrow^m (n-1)) < 3 \uparrow^{m-1}(3\uparrow^m (n-1)) = 3\uparrow^m n$$

For the right inequality, we have

$$2 \uparrow^m (n+2)-2 = 2\uparrow^{m-1} (2\uparrow^m (n+1)) - 2 > 2\uparrow^{m-1}(3 \uparrow^m (n-1)+2) - 2 > 3\uparrow^{m-1}(3\uparrow^m (n-1)) = 3\uparrow^m n$$

And we're done.