A change of thinking

We need to consider the function $f(n) \; = \; \left\lceil 1000\sum_{p=2}^n \zeta(p) \right\rceil - 1000x \;, \qquad \qquad n \ge 2 \;,$ and calculate that $f(n)$ is equal to $-355,-153, -70,-33, -16,-8, -3,-1,0$ for $n \,=\, 2,3,\ldots,10$ . The answer is $10$ .

Of course, since $\sum_{p=2}^\infty \big(\zeta(p) - 1\big) \; = \; 1 \;,$ we can show that $f(n) \; = \; -\left\lfloor 1000\sum_{p > n} \big(\zeta(p) - 1\big) \right\rfloor \;, \qquad \qquad n \ge 2 \;,$ which means that the sequence $\{f(n)\}_{n \ge 2}$ is an increasing sequence of integers tending to $0$ , and hence there exists an integer $N$ such that $f(n) < 0$ for all $n < N$ , with $f(n) = 0$ for all $n \ge N$ . Finding this value of $N$ is the same as asking for the smallest value of $n$ such that $\sum_{p=2}^n \big(\zeta(p) - 1\big) \; > \; 0.999 \;.$