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In this question we determine that the series $\sum_{p \in \mathcal P} \frac1{p\ln p}$ converges, where the sum runs over primes.

As I see the convergence is really slow. The partial sums for given $N$ finite upper limits are

$\begin{align} N \quad & \text{partial sum}\\ 100 \quad & 0.757042464018193\\ 1000 \quad & 0.803993788114564\\ 10000 \quad & 0.828779261095689\\ 100000 \quad & 0.844238045700797\\ 1000000 \quad & 0.854866046633956\\ \end{align}$

Upto the $1000000$th partial sum there is no significant digit. Could anyone give me the sum of this series for some significant digits? As many as you can, but at least $10$ digits would be nice.

Edit. My calculations above have an $1/(2 \ln 2)$ difference, because the sum runs from $p_2$.

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user153012
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  • I get partial sum $1.421567...$ for $N=97$ – Peter Nov 19 '14 at 20:31
  • For the first 10,000 primes, I get $1.5501267...$ – Simon S Nov 19 '14 at 20:34
  • You are abolutelty write, in my calculations the sum runs from $p_2$. – user153012 Nov 19 '14 at 20:35
  • You sum looks wrong, the sum to $1000000^{th}$ prime is around $1.57621356707841$.

    In July 1998, Henri Cohen has computed the limit with up to 50 digits accuracy. $$\sum_{p \in \mathcal{P}} \frac{1}{p\log p} \sim 1.63661632335126086856965800392186367118159707613129\ldots$$

     – achille hui Nov 19 '14 at 20:37
  • @achillehui Yes, sorry. My calculation has an $1/2$ difference, because the sum runs from $p_2$. Could you give me a reference to the result what you've said? – user153012 Nov 19 '14 at 20:39
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    Your difference should be $1/(2 \log 2)$, not $1/2$. – Michael Lugo Nov 19 '14 at 20:43
  • All the source I have refer to a paper "High precision computation of Hardy-Littlewood constants" by Henri Cohen. On Cohen's homepage, there is a draft for the preprint. But it is in dvi format which I cannot view. If you can view dvi file, give that a try. – achille hui Nov 19 '14 at 20:45
  • $$998200001 \ \ \ \ 1.588357254141502075976500372$$ – Peter Nov 19 '14 at 20:46
  • So, even the primes upto $998200001$ lead to a very inaccurate result. – Peter Nov 19 '14 at 20:46
  • ? s=0;p=1;while(p<10^9,p=nextprime(p+1);s=s+1/p/log(p);if(Mod(p,10^5)==1,print(p ," ",s))) – Peter Nov 19 '14 at 20:48
  • I used the above command in PARI/GP to get the result. – Peter Nov 19 '14 at 20:48
  • The final result is :? print(p," ",s) $1000000007 \ \ \ \ 1.588361447267763827936188124$ – Peter Nov 19 '14 at 20:49
  • So, the convergence is very slow indeed, which is no wonder because $$\sum_p \frac{1}{p}$$ is still divergent. – Peter Nov 19 '14 at 20:51
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    From the Henri Cohen paper: "Thus, it would be inconceivable to compute it using the naive method since even with a table of primes up to $10^{20}$ (already an impossible practical limit), we would obtain less than 2 decimal digits." – vadim123 Nov 19 '14 at 20:52
  • Wow, very very slow ... – Peter Nov 19 '14 at 20:53
  • Sorry about the mistakes, and thanks for the helpful comments. Yes, really slow. – user153012 Nov 19 '14 at 21:02
  • I have previously asked a similar question at http://math.stackexchange.com/q/888571/131263. The answer there might be useful. – barak manos Nov 19 '14 at 21:20

1 Answers1

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Your sums seem wrong; they are all off by an additive factor of 1/(2 log 2). Why you omitted the prime 2 is unknown to me. Also by N you seem to mean "compute the sum up to and including the N'th prime".

In any event, you can find the value of this sum to about 45 digits here:

https://oeis.org/A137245

Jeffrey Shallit
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