I found exercise in my book for number theory that I can't resolve. How do you show that $$p_n < e^{1+n}$$ where $p_n$ is $n$-th prime number?

  • $\begingroup$ I think this follows from Bertrand's Postulate...are you allowed to assume that? $\endgroup$
    – Nishant
    Sep 6, 2014 at 23:08
  • $\begingroup$ Bertrand's postulate is proofed, so We can use it :) But the best is the easiest proof. $\endgroup$ Sep 6, 2014 at 23:17
  • $\begingroup$ There are many results about the bounds of the $n$th prime. They perhaps are helpful, chee up :) $\endgroup$
    – Yes
    Sep 6, 2014 at 23:30
  • $\begingroup$ @TheKwiatek666 So you want a proof that does not use Bertrand's Postulate? $\endgroup$
    – graydad
    Sep 7, 2014 at 0:08

1 Answer 1


Let's try induction. For the base case of $n=1$, it is clear that $p_{1}=2<e^{1+1}=e^2$. Now let's suppose for all $k$ where $n \geq k >1$ that our result holds. We know by Bertrand's Postulate that $p_{k+1}<2p_{k}$ for all $k \in \mathbb{N}$ and by induction we know $2p_{k}<2e^{k+1}$. It follows that $$p_{k+1}<2e^{k+1}<(e)e^{k+1}=e^{k+2}$$ We now know our result holds for all $k \in \mathbb{N}$.

  • $\begingroup$ This is one of the first exercises in this book. Maybe is it possible to resolve that without Bertrand's Postulate? This theorem is proofed four chapters later. $\endgroup$ Sep 7, 2014 at 14:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .