# Sequence growing slower than polynomial has infinitely many prime divisors

Suppose we have a monotonic sequence $$a_n$$ such that $$a_n \leq f(n)$$ for some polynomial $$f$$ then there are infinitely many primes that divide some $$a_k$$.

I only know how I can prove this for linear polynomials. Let $$a_n \leq an+b \forall n\geq 1$$ Suppose that $$p_1,p_2,\ldots,p_m$$ be the primes dividing $$a_k$$ then : $$\sum_{j=1}^{\infty}{\dfrac{1}{aj+b}} \leq \sum_{j=1}^{\infty}{\dfrac{1}{a_j}} \leq \prod_{j=1}^{m}{\sum_{k=0}^{\infty}{\dfrac{1}{p_j ^k}}}$$ The right sum is finite by geometrical series but for sufficiently large $$\alpha$$ , $$\sum_{j=1}^{\infty}{\dfrac{1}{aj+b}}\geq \dfrac{1}{a+\alpha} \sum_{j=1}^{\infty}{\dfrac{1}{j}}$$ diverges.

Actually a similar approach works: Let $$k=\frac{1}{\text{2deg(f)}}$$ then we compare $$\sum_{i=1}^{\infty}{\dfrac{1}{f(i) ^k}}\leq\sum_{i=1}^{\infty}{\dfrac{1}{a_i ^k}}\leq\prod_{j=1}^{m}{\sum_{i=0}^{\infty}{\dfrac{1}{p_j ^{ik}}}}$$ Then by GP RHS is finite and LHS is infinite. $$\square$$