Does the sum of the inverses of the sums of the primes converge?

All you need is $p_n \ge n$, so $\sum_{n=1}^m p_n \ge \sum_{n=1}^m n = m(m+1)/2$, and $\sum_{m=1}^\infty \dfrac{2}{m(m+1)}$ converges.

The answer is yes.

For large $m$, the denominators are known to be approximately $\frac{1}{2}m^2\ln m$; in particular, they are larger than $m^2$. Hence for $m$ sufficiently large, the terms are bounded above by $\frac{1}{m^2}$.

The n-th prime is quite obviously ≥ 2n - 1. Therefore the sum of the first n primes is ≥ $n^2$. The sum in question converges and is at most the sum over $1 / n^2$ which converges to $π^2 / 6$. Actually, it must be less than $π^2 / 6 - 1/2$ because the first element of the sum is 1/2, not 1.

That sum is about 1.1493, not worlds apart from the actual result.

I'm just curious why an answer saying "all you need is $p_n ≥ n$" gets voted up while an answer saying "$p_n ≥ 2n - 1$" and giving a reasonable upper bound for the limit doesn't.