0
$\begingroup$

Research done:
There are 4 prime numbers between 0 and 10. This means there are 4 primes whose reciprocals are > 0.1. The exact sum of these prime numbers = $\frac{247}{210}$ $\approx 1.176$

There are 21 prime numbers between 10 and $10^2$. This means there are 21 prime numbers whose reciprocals are between $10^{-1}$ and $10^{-2}$ The sum of the reciprocals of these prime numbers is $\approx 0.6266$
There are 143 primes between $10^2$ and $10^3$. This means that there are 143 primes whose reciprocals are between $10^{-2}$ and $10^{-3}$.The sum of the reciprocals of these prime numbers is $\approx \frac{143}{525}$ $\approx 0.2724$
There are 1061 primes between $10^3$ and $10^4$. This means that there are 1061 primes whose reciprocals are between $10^{-3}$ and $10^{-4}$. The sum of the reciprocals of these primes is $\approx \frac{1061}{5335} \approx 0.19888$
There are 8363 primes between $10^4$ and $10^5$. This means that there are 8363 primes whose reciprocals are between $10^{-4}$ and $10^{-5}$. The sum of the reciprocals of these primes is $\approx \frac{8363}{53648} \approx 0.155886$
There are 68906 primes between $10^5$ and $10^6$. This means that there are 68906 primes whose reciprocals are between $10^{-5}$ and $10^{-6}$. The sum of the reciprocals of these primes $\approx \frac{68906}{538357} \approx 0.12799$.
There are 586081 primes between $10^6$ and $10^7$. This means that there are 586081 primes whose reciprocals are between $10^{-6}$ and $10^{-7}$. The sum of the reciprocals of these primes $\approx \frac{586081}{5401599} \approx 0.1085$.

(The calculations have been done with a high accuracy only not completely shown)

Question: As we can clearly see the increase of the sum diminishes as prime numbers get higher and higher. However, this still remains to be proved for all primes. However if this would be true: you would get the sum of the series: 1.176+0.6266+0.2724+0.19888+0.155886+0.12799+0.1085... where the values keep getting lower. Does such a series necessarily diverge to infinity?

$\endgroup$
0
$\begingroup$

Yes, diverges, proved by Euler in early 18th century: https://en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes

$\endgroup$
  • $\begingroup$ I know Euler's proof. He arrives at it from a very different angle. The question at hand is: if you have the sum of a series of numbers which diminishes do you get a divergent series? This might be a different question than Euler posed. Please also check: math.stackexchange.com/questions/1492372/… $\endgroup$ – St.Clair Bij Oct 22 '15 at 16:53
  • $\begingroup$ (You must acknowledge that it is a bit weird that the sum of the reciprocals of the primes to 10 million equals $\approx$ 2.666556 and less and less will be added when you take prime numbers between 10 million and 100 million, and 100 million and 1 billion.) $\endgroup$ – St.Clair Bij Oct 22 '15 at 17:02
  • $\begingroup$ "if you have the sum of a series of numbers which diminishes do you get a divergent series?" What does this mean? For any convergent sum $\sum a_n$, it must be true that $a_n\to 0$. $\endgroup$ – Rocket Man Oct 22 '15 at 17:04
  • $\begingroup$ @St. Clair Bij: You choose a very very specific delimiters for grouping - if you want the series to be convergent, diminishing of partial sums should occur for any grouping method you can imagine. $\endgroup$ – z100 Oct 22 '15 at 17:19
  • 1
    $\begingroup$ maybe it is possible to construct group delimiters based on the idea as Paul Erdos's proof by contradiction (see in upper link) $\endgroup$ – z100 Oct 22 '15 at 17:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.