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$$\sum_{m=0}^∞ \frac{1}{\sum_{n=0}^m p_n} = \frac{1}{2} + \frac{1}{5} + \frac{1}{10} + \frac{1}{17} ... $$

Where $p_n$ is the $n$th prime number, does $\sum_{m=0}^∞ \frac{1}{\sum_{n=0}^m p_n}$ converge?

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The sum of the first $n$ primes is asymptotic to $\frac{1}{2}n^2 \ln n$, so yes. – Slade Feb 26 at 4:01
The value of the sum is OEIS A122989, namely $1.02347632392012\ldots$. – Jeppe Stig Nielsen Feb 26 at 12:43
up vote 61 down vote accepted

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.

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That is by far the simplest answer. +1 – Patrick Da Silva Feb 26 at 16:42
(+1) in fact, your sum converges to $2$. – robjohn Feb 26 at 16:52

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 $\frac{1}{m^2}$.

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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.

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Whining about votes comes off as very unbecoming... – Zach466920 Feb 28 at 1:35

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