If we have a sequence $\left\{\alpha_{n}\right\}_{n=3}^{\infty}$
such that $\alpha_{n}$ is the least prime divisors of $n !-1$
To Show: $$\sum_{n=3}^{\infty} \frac{1}{\alpha_{n}} \rightarrow \infty$$
I need help in completing my proof :
$\Rightarrow$ Claim : The least prime divisor of $n!-1$ is greater than $n$.
If on the contrary we have prime $p$ s.t. $p \leq n$
then clearly $p \mid n !$
and if we assume $p$ also divides $n !-1$ then $p$ divides $n !-(n !-1)=1$
Hence contradiction.
So the least prime divisor of $n !-1$ is greater than $n$.
Now how to prove that $\sum_{n=3}^{\infty} \frac{1}{\alpha_{n}} \rightarrow \infty$ once I have shown $\alpha_n \in \{n+1, \ldots, n !-1 \}$.
Best Answer
Wilson's theorem helps to prove that the sequence diverges : For all integers $n>1$ for which $n+2$ is prime , we have $$(n+2)\mid n!-1$$
This follows from $$(n+1)!\equiv -1\mod (n+2)$$ and $$n+1\equiv -1\mod (n+2)$$
In this case, the smallest prime factor is obviously $n+2$
Hence the sum contains all the reciprocals of the primes $p\ge 5$. Since the sum of the reciprocals of the primes diverges, the claim follows.