I am interested in the following sum
\begin{equation} \sum_{i=1}^m\left\{\frac{1}{p_i}\prod_{j=i+1}^m\left(1-\frac{1}{p_j}\right)\right\}, \end{equation} where $p_i$ is the i'th prime number.
I assume that finding a closed form expression for any $m$ is very difficult, so I am interested in approximations or upper and lower bounds on these sums. I do think this sum converges for $m\rightarrow\infty$ but I am not sure.
An obvious upper bound is just $\sum_{i=1}^m\frac{1}{p_i}$, but it is definitely not a good one.
Since the factors in the product can be lower bounded by $(1-\frac{1}{j})$, the sum can be lower bounded by $\sum\frac{1}{p_i}\frac{i}{m}$. But this is also not a good lower bound.
edit: I forgot to mention that $p_i$ represents the i'th prime number.
Obviously,$$\frac{1}{p_i}\prod_{j=i+1}^m\left(1-\frac{1}{p_j}\right)=\prod_{j=i+1}^m\left(1-\frac{1}{p_j}\right)-\prod_{j=i}^m\left(1-\frac{1}{p_j}\right),$$ because $$\prod_{j=i}^m\left(1-\frac{1}{p_j}\right)=\left(1-\frac{1}{p_i}\right)\,\prod_{j=i+1}^m\left(1-\frac{1}{p_j}\right),$$ so the telescoping sum will be $$\sum_{i=1}^m\left\{\frac{1}{p_i}\prod_{j=i+1}^m\left(1-\frac{1}{p_j}\right)\right\}=1-\prod_{j=1}^m\left(1-\frac{1}{p_j}\right).$$