15
$\begingroup$

I'm looking to prove that:

$$p_n \leq 2^{2^{n-1}}$$

Where $p_n$ denotes the $n$th prime in ascending order. The proof method is induction. I've solved my base case, that is: $n=1$ $p_1 = 2$, $2^{2^0}=2$, $2\leq2$ Therefore $P(1)$ is true.

And the induction hypothesis is that $P(1), P(2),\ldots P(k)$ is true for some integer $k$.

I'm stuck trying to prove $P(k+1)$, or $p_{k+1} \leq 2^{2^k}$

The closest I have come was rewriting $2^{2^k}$ as $(2^{2^{k-1}})^2$, which is the square of the value of $P(k)$. My guess at the moment is that if I can prove that a prime is strictly less then the square of the previous prime then my proof would be complete, but I can't seem to do so.

Thanks in advance for help.

  • 4
    +1 for showing your thoughts about the problem - you are an exemplary new user! Welcome to math.SE!2011-11-16
  • 0
    More -and similar- answers may be found at the much related question from about two monthes ago at MSE: http://math.stackexchange.com/questions/656302011-11-16
  • 0
    Is there a thing as primes in descending order.2011-11-16

2 Answers 2