TITLE: Does the sum of the inverses of the sums of the primes converge?
QUESTION [36 upvotes]: $$\sum_{m=0}^∞ \frac{1}{\sum_{n=0}^m p_n} = \frac{1}{2} + \frac{1}{5} + \frac{1}{10} + \frac{1}{17} ... $$
Where $p_n$ is the $n$th prime number, does $\sum_{m=0}^∞ \frac{1}{\sum_{n=0}^m p_n}$ converge?

REPLY [4 votes]: 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.