# Help with an interesting reciprocal prime sum

How can you evaluate $$\sum_{n = 1}^\infty \left( \frac{1}{\sum (n)} \right)$$ where $$\sum (n)$$ is the sum of the first $$n$$ prime numbers? I was able to find numerical bounds by running code to give a lower bound of $$1.02345$$, and I found an upper bound by taking the reciprocal of the Prime Sum Function approximation $$\frac{\ln(n)\cdot n^2}{2}$$ and evaluating a sum from there, but I would like to figure out how to evaluate this exactly.

• Most randomly chosen infinite series cannot be evaluated in closed form. I don't see any reason why there would be a closed form for this series. Nov 23 '20 at 8:47
• Thank you, that is very interesting. However, as this sum isn't random and sums the primes, shouldn't there be a closed form? I'm curious:) Nov 23 '20 at 9:05
• Most sums don't have closed form. Nov 23 '20 at 9:43

You want to compute $$S=\sum_{n = 1}^\infty \frac{1}{a_n} \qquad \text{where} \qquad a_n=\sum _{i=1}^n p_i$$ What I should do it to write it as $$S=\sum_{n = 1}^p \frac{1}{a_n}+\sum_{n = p+1}^\infty \frac{1}{a_n}$$
Concerning $$s_p=\sum_{n = 1}^p \frac{1}{a_n}$$ they form the sequence $$\left\{\frac{1}{2},\frac{7}{10},\frac{4}{5},\frac{73}{85},\frac{2129}{2380},\frac {89669}{97580},\frac{2649191}{2829820},\frac{29545361}{31128020},\frac{7464160 3}{77820050}\right\}$$ For the second summation, I should use as an approximation of $$a_n$$ $$\frac{n^2}{2}\left[\ln n + \ln\ln n - \frac{3}{2} + \frac{\ln\ln n}{\ln n} - \frac{5}{2\ln n} - \frac{\ln^2\ln n}{2\ln^2 n} + \frac{7\ln\ln n}{2\ln^2 n} - \frac{29}{4\ln^2 n} + \cdots\right]$$ as proposed by Sinha in $$2011$$.