Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $P(k)$ be the product of $k$ consecutive primes $p_1, p_2, \dots, p_k$. So, e.g. $P(4)$ is $2 \cdot 3 \cdot 5 \cdot 7 = 210$.

Is anything known about whether $P(k) > p_{k+1}$ is always true (for $k > 1$)? It seems like it should be true, since $P(k)$ gets large quickly (faster than $k!$), but I'm having trouble seeing how to construct a counterexample.

If there were a counterexample, then it seems like this would violate PNT locally if it were true since the primes would have to be very sparse in that region. But that's not definitive. Is there an elementary proof one way or another?

share|cite|improve this question

By Bertrand's Postulate, which has been a theorem for a long time, there is always a prime between $n$ and $2n$.

share|cite|improve this answer

Look at the integer $n=P(k)-1=p_1p_2\dots p_k - 1$.
Here $n>p_k$, as long as we look for cases for $k>1$.

If $n$ is prime, then certainly $p_k<p_{k+1}\leq n$.
Then we are done since $P(k)=n+1\implies P(k)>n\geq p_{k+1}\implies P(k)>p_{k+1}$.

Hence suppose $n$ is a composite.
Let $r$ be any divisor of $n$, then $n\equiv 0(\text{mod }r)$.
Since $n\equiv-1\not\equiv 0(\text{mod }p_i)$, we conclude that $r\not\in\lbrace p_1,\dots,p_k\rbrace$.
i.e. there is a prime $p_k<r<n$.

Since $n$ is finite, we may find a minimal $r$.
This $r$ must be $p_{k+1}$, which certainly satisfies $p_{k+1}<n<P(k)$.
Therefore we reach the conclusion that $P(k)>p_{k+1}$.
i.e. the statement is always true.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.