The denominator is divisible by all primes $p$ such that $\dfrac{n}{\log{n}} << \ll p < n$, so the denominator grows exponentially in $n$. Furthermore, I believe (and have been too lazy and incompetent to check) that the sum is (in absolute value) $>> \gg n^{-t}$. This implies that the numerator also grows exponentially in $n$. So to me this shows that the problem should be easy; how can there ever not be a prime $> cn$ dividing somethig like $e^n$? This is espeically true for $c=1$, since if we know that the numerator is only divisible by primes smaller than $n$, we actually know that all the prime divisors of it must be $<< \ll \dfrac{n}{\log{n}}$. So, since Gerry pointed out that my metaconjecture has been refuted, I'd like to coin a new one: for $c=1$ the conjecture is provable by a simple counting argument.
The denominator is divisible by all primes $p$ such that $\dfrac{n}{\log{n}} << p < n$, so the denominator grows exponentially in $n$. Furthermore, I believe (and have been too lazy and incompetent to check) that the sum is (in absolute value) $>> n^{-t}$. This implies that the numerator also grows exponentially in $n$. So to me this shows that the problem should be easy; how can there ever not be a prime $> cn$ dividing somethig like $e^n$? This is espeically true for $c=1$, since if we know that the numerator is only divisible by primes smaller than $n$, we actually know that all the prime divisors of it must be $<< \dfrac{n}{\log{n}}$. So, since Gerry pointed out that my metaconjecture has been refuted, I'd like to coin a new one: for $c=1$ the conjecture is provable by a simple counting argument.