prime number question
- does the series 1/p, where p are prime numbers, converge or diverge?
please cc the answer to ocsurfer19@...
keep up the good work,
- Diverges. If you take all the primes p<=x, and sum 1/p, the value is
loglog x+C, where C is some number which tends to a fixed value as x
goes to infinity.
This formula, and others, are elementary properties of the series of
primes, and can be found in practically any basic number theory book.
> does the series 1/p, where p are prime numbers, converge or diverge?
> please cc the answer to ocsurfer19@...
> keep up the good work,
- --On Thursday, August 07, 2003 1:21 AM -0700 Paul Leyland
>> From: ansamanta [mailto:ocsurfer19@...]
>> does the series 1/p, where p are prime numbers, converge or
>>number of primes.
>> please cc the answer to ocsurfer19@...
>> keep up the good work,
> It diverges, though very slowly.
> This fact alone provides a proof that there are an infinite
>And that the proportion of numbers that are prime decreases more
slowly than n^m, where m < -1.
I vaguely remember that for m < -1, the sum is approximated by
(picture an integral sign instead of an "S")
S n^m dn = [ 1/m * n^(m-1) ]
(note that at m=-1 this is a log instead, and log(oo)= oo , so
it diverges, and for 0>m>-1 it diverges since each term is
greater than the corresponding term with m = -1, I'm not sure
how to justify it in terms of the calculus though).
Anyhow, we have a converging series for all m < -1, which means
that the primes decrease in density at a speed only slightly
less than linear.
Sorry for rambling, it's been 3 years since I took any formal
math besides boolean logic and related fields.