Sum and products of consecutive integers

$\dfrac{1 \times 2 \times \cdots \times m}{1 + 2 + \cdots + m} = \dfrac{m!}{\dfrac{m(m + 1)}{2}} = \dfrac{2(m - 1)!}{m + 1}$ .

Now for any even positive integer $m$ such that $m + 1$ is prime, $2(m - 1)!$ will not have $m + 1$ as a term in its prime factorization, i.e., will not be divisible by $m + 1$ . So since there are an infinite number of primes, there will be infinitely many such integers $m$ .

$\dfrac{1\times 2\times 3\times \cdots \times k}{1+ 2+\cdots +k} = \dfrac{2(k-1)!}{k+1}$ .Note that $k$ is an even number so $k-1$ and $k+1$ are odd integers where $k+1 > k-1$ which shows that $(k-1)$ has no factor of $k+1$ .

Further note that $2$ and $k+1$ are co-prime to each and difference between $k-1$ and $k+1$ is of $2$ so easily we concluded that if $k-1$ is prime then $k+1$ will be an odd number and vice or both will be prime. Hence there are infintely many even numbers. $m$

Similar problem here .