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 $p_2$.

  • $\begingroup$ I get partial sum $1.421567...$ for $N=97$ $\endgroup$ – Peter Nov 19 '14 at 20:31
  • $\begingroup$ For the first 10,000 primes, I get $1.5501267...$ $\endgroup$ – Simon S Nov 19 '14 at 20:34
  • $\begingroup$ You are abolutelty write, in my calculations the sum runs from $p_2$. $\endgroup$ – user153012 Nov 19 '14 at 20:35
  • 1
    $\begingroup$ Your difference should be $1/(2 \log 2)$, not $1/2$. $\endgroup$ – Michael Lugo Nov 19 '14 at 20:43
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    $\begingroup$ 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." $\endgroup$ – vadim123 Nov 19 '14 at 20:52

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:



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