Im trying to show that there exist real numbers $b$ and $C$ such that $|b-\sum_{p<n}\log(1-\frac{1}{p})+\frac{1}{p}|<\frac{C}{n}$ for any positive integers $n$, where $p$ represents a prime number.
So far, I've tried to find an upper bound for the sum, which I'll denote by $S_{n}$. I have that $\sum_{p<n}\log(1-\frac{1}{p}) \leq 0$ and also that $\sum_{p<n} < \sum_{k=1}^{n} = \log(n)+\gamma+O(\frac{1}{n})$. Combining this we get that $S_{n} <\log(n)+ \gamma + O(\frac{1}{n})$ but I'm not sure what to conclude from this. Is there something worth pursuing here?
My second though was to use this this to show that $S_{n}$ converges, then set $b=\lim S_{n}$. Then we can pick $\varepsilon > 0$ and get that $|b-S_{n}| < \varepsilon$ for all $n$ past some certain $N$, so setting $C = max_{1 \leq j \leq N}\{|b-S_{n}|+\varepsilon \}$ we have $|b-S_{n}|<C$ for all $n$. While this isn't the solution I feel like it might be moving in the right direction. Can I do better than this using convergence?