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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?

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2 Answers 2

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

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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.

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