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Let $P_n$ be the $n$th prime. Ie $P_1 = 2, P_2 = 3, P_3 = 5, ...$ Show that $P_n \le 2^{2^n}$

Induction seems to be useless here. Is there a hint someone can provide?

hhhhhhhajajaja
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    https://math.stackexchange.com/questions/65630/how-to-show-p-n-leq-22n?rq=1 – Marco Lecci Jun 18 '17 at 12:46
  • The set of square-free numbers $E$ has a positive asymptotic density, $\frac{1}{\zeta(2)}=\frac{6}{\pi^2}$. Let $q=p_n$ be the $n$-th prime: every element of $E\cap[1,q)$ can be written as a product of primes $<q$, hence $$ 2^{n-1}+1 \geq \frac{6}{\pi^2} p_n$$ has to hold for any $n$ large enough. This leads to the improved inequality $$ p_n \leq \color{red}{\frac{5}{6},2^n}.$$ – Jack D'Aurizio Jun 18 '17 at 13:51

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Use induction. For the inductive step, $$P_{n+1} \leq \left( \prod_{i=1}^n P_i \right)+1 \leq \left(\prod_{i=1}^n 2^{2^i}\right)+1 = 2^{\sum_{i=1}^n2^i}+1$$ $$= 2^{2^{n+1}-2}+1 \leq 2^{2^{n+1}}$$ The first inequality comes from the fact that $P_1\cdot P_2\cdot ...\cdot P_n+1$ shares no common factors with $P_1,P_2,...,P_n$, implying it has some prime greater than or equal to $P_{n+1}$ as a factor.

florence
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  • Nice and elegant solution (+1) – BAI Jun 18 '17 at 13:22
  • This method can be used 'as is' to show that $P_n\le2^{2^{n-1}}$. – TonyK Jun 18 '17 at 13:33
  • How does $$\gcd(P_1...P_n, P_1....P_n + 1) = 1$$ imply that $$P_1...P_n + 1 \ge P_{n+1}$$? – hhhhhhhajajaja Jun 18 '17 at 21:07
  • Let $x=P_1\cdot P_2\cdot...\cdot P_n+1$. Let $p$ be a prime factor of $x$. Then $p=P_m$ for some $m$. Since $P_1$ through $P_n$ are the first $n$ primes, and since none of $P_1,...,P_n$ divide $x$, we have $m\geq n+1$. Thus, $P_{n+1}\leq P_m \leq x$. The last inequality is true since $a\vert b \implies a\leq b$. – florence Jun 18 '17 at 23:17