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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.

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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. $\square$

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