Does anyone know of any information/work on this sum? I found absolutely nothing on the web about it.

3$\begingroup$ Naive question: why does one expect the value of this sum to have interesting properties? $\endgroup$– Yemon ChoiJan 27, 2011 at 3:47

5$\begingroup$ Ah, I see that someone more learned than me has provided links below, which presumably answer my question $\endgroup$– Yemon ChoiJan 27, 2011 at 3:49

2$\begingroup$ Kudos for not deleting "why does one expect the value of this sum to have interesting properties?" In retrospect, it was extremely degrading to Ethan Brush, wasn't it?  Well, he never came back to this site. $\endgroup$– nilo de roockMar 8, 2021 at 18:16

$\begingroup$ @niloderoock Your definition of "extremely degrading" doesn't agree with mine. $\endgroup$– Robert FurberJan 20 at 20:17

$\begingroup$ Something must have been deleted because I don't understand my own reaction from what I can see here. $\endgroup$– nilo de roockJan 28 at 11:12
1 Answer
This would be $P(2)$, where $P$ is the "prime zeta function," q.v.
A couple of very old references are C. W. Merrifield, The Sums of the Series of Reciprocals of the Prime Numbers and of Their Powers, Proc. Roy. Soc. London 33 (1881) 4–10. doi:10.1098/rspl.1881.0063. JSTOR 113877
J. W. L. Glaisher, On the Sums of Inverse Powers of the Prime Numbers, Quart. J. Math. 25 (1891) 347–362.
EDIT: A more recent source is Steven R Finch, Mathematical Constants, page 95: The sum of the squared reciprocals of primes is $$N=\sum_p{1\over p^2}=\sum_{k=1}^{\infty}{\mu(k)\over k}\log(\zeta(2k))=0.4522474200\dots$$

66$\begingroup$ suspiciously close to the lb/kg ratio.... $\endgroup$ Jan 27, 2011 at 4:18

1$\begingroup$ An older source is in Euler's Introductio in analysin infinitorum, volume 1 chapter 15 (1748) eulerarchive.maa.org/pages/E101.html . An English translation is available here 17centurymaths.com/contents/introductiontoanalysisvol1.htm $\endgroup$ Jun 11, 2016 at 9:50

1$\begingroup$ In particular, on page 480 of 17centurymaths.com/contents/euler/introductiontoanalysisvolone/… $\endgroup$ Jun 11, 2016 at 14:02

$\begingroup$ hahaha, @YaakovBaruch nice; $4/(\pi^2 1)$ is close too hahaha $\endgroup$ Apr 12, 2017 at 20:43