Elementary Methods in Number Theory Exercise 1.5.13

Let $2=p_1<p_2<\cdots$ be the sequence of primes in increasingorder.Prove that
\begin{equation}
p_n\leq 2^{2^{n-1}}
\end{equation}for all $n\geq 1$.

Proof:When $n=1$,$2\leq 2$.Suppose $\forall n\leq k$,
\begin{equation}
p_n\leq 2^{2^{n-1}}
\end{equation}
Then
\begin{equation}
p_{n+1}\leq p_1p_2\cdots
p_n+1\leq 2^{1+2^1+\cdots+2^{n-1}}+1(p_1p_2\cdots p_n+1~\mbox{is a prime.})=2^{2^n-1}+1\leq 2^{2^n}
\end{equation}
By induction,$\forall n\in\mathbf{N}^{+}$,
\begin{equation}
p_n\leq 2^{2^{n-1}}
\end{equation}

转载于:https://www.cnblogs.com/yeluqing/archive/2012/12/02/3828020.html