Let $a_{1}=2$, $a_{n+1}=2^{a_{n}}$ for $n \geq 1$

Let $b_{1}=3$, $b_{n+1}=3^{b_{n}}$ for $n \geq 1$

Is is true that $a_{n+2}>b_{n}$ for all $n \geq 1$?

If so, is the proof elementary? (Use only Mathematical techniques covered in an undergraduate degree)


Yes. Here is a simple proof by induction:

Theorem. $a_{n+2} > 4 b_n$

If $n=1$, then $a_3 = 16 > 4*3 = 4b_1$.

Now assume $a_{n+2} > 4 b_n$:

$$a_{n+3} =2^{a_{n+2}} > 2^{4b_n} = 16^{b_n} > 4^{b_n} * 3^{b_n} > 4 * 3^{b_n} = 4b_{n+1}$$

and the result follows.

  • $\begingroup$ Thanks, very nice solution indeed. That is a clever way of using mathematical induction. I try to induct using $a_{n+2}>b_{n}$ previously but fail to make any progress. $\endgroup$ – ZK Lin Feb 13 '16 at 0:24

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