Finding an upper bound for a sum over primes

Question

Fix $X>\geq 1$ a real number and let $1\leq y<X.$ Is there a positive constant $B$ such that

$$\prod_{y<p\leq X} \left(1+\frac{3}{p}+ \sum_{\nu \geq 2} \frac{(\nu+1)^2}{p^{\nu}}\right)\leq B.$$ Thank you!

Answer 1

To find a $B$ that works for all $X$ is impossible.

There is a standard result that, if $a_n$ is a sequence with $\sum_n |a_n|^2 < \infty$, then convergence of $\prod_n (1+a_n)$ is equivalent to convergence of $\sum_n a_n$.

Let $a_n = \frac{3}{p_n} + \sum_\nu \frac{(\nu+1)^2}{p_n^\nu} = \frac{3}{p_n} + O(p_n^{-2})$. Then $\sum_n |a_n|^2$ converges, so we can apply the result.

But the sum of the reciprocals of the primes diverges, so the product diverges as well.