# Prove or disprove that $p_n > e^{p_n - p_{n-1}}$ for large enough $n$.

Let $p_n$ denote the $n$-th prime. Prove or disprove that for large enough $n$ we have $$p_n > e^{p_n - p_{n-1}}.$$

The inequality trivially holds for all the twin primes larger than $7$. With $n$ larger than some $N$, does it always hold? I think not, but don't really know how to prove it.

## 2 Answers

A disproof: Rankin showed that there are infinitely many $n$ for which $$p_n-p_{n-1} > \log n (\log\log n)^{1/2}$$ (in fact his result was rather stronger than this). In particular, there are infinitely many $n$ such that $$p_n-p_{n-1} > 2\log n > \log p_n$$ (the latter inequality is just asserting that $p_n < n^2$ for sufficiently large $n$). Exponentiation both sides yields $$e^{p_n-p_{n-1}} > p_n$$ for infinitely many $n$.

When one exponentiates an expression (like $p_n-p_{n-1}$), a tiny difference between it being of size $0.99\log n$ and $1.01\log n$ gets magnified so much that it makes all the difference in the world compared to $p_n$.

• Thank you. I was about to accept that first invalid proof when I realized, indeed, there was probably something wrong and deleted my comment at it. – Vincenzo Oliva Dec 5 '14 at 9:05

Rankin showed that for infinitely many values of n,

$$g_n > \frac{c\log n\log^2 n\log^4 n}{(\log^3 n)^2}.$$

Then $$e^{g_n}> n^{\frac{c\log^2 n\log^4 n}{(\log^3 n)^2}} = n^{f(n)}$$

$f(n)$ eventually exceeds 2 so

$$e^{g_n} > p_n$$ for infinitely many values of n.

• Apologies, I'm at school at the moment so I didn't have the time to upvote your answer until now that there's the interval. (but I still think accepting an answer is already a kind of upvote, so I usually upvote only other answers)Thank you too! – Vincenzo Oliva Dec 5 '14 at 10:12
• @VincenzoOliva: Yes, it's true that accepting confers points, and no problem--I am always curious about how people vote so it was really just a question. Thanks. – daniel Dec 5 '14 at 10:14