Reciprocals of Prime-Primes
Let P k be the k th prime: P 1 = 2 , P 2 = 3 , and so on. As it turns out, k = 1 ∑ ∞ P k 1 = 2 1 + 3 1 + 5 1 + 7 1 + ⋯ diverges.
Does k = 1 ∑ ∞ P P k 1 = 3 1 + 5 1 + 1 1 1 + 1 7 1 + ⋯ converge?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
Moderator note:
Nice solution using the comparison to a "standard" sum. It would be helpful to explain how exactly you went from the logic of the approximations to the strict inequalities (for sufficiently large k ).
A natural extension to this question is to describe all S ⊂ N such that ∑ s ∈ S P s 1 converges.
But what does it converge to? :)
Log in to reply
I doubt that this rather contrived sum can be evaluated in closed form.
It's not needed but according to Rosser's theorem it's impressively true that P k > k ln ( k ) for k ∈ N .
Log in to reply
Interesting! Thanks!
Log in to reply
I'm your 700th follower :)
Log in to reply
@Aryan Gaikwad – I'm honoured... I'm your 120th follower now!
Might there be some way to prove this that does not rely on something whose proof is as onerous as that of the prime number theorem? The divergence of the sum of the reciprocals of the primes is far easier to prove than that, as is the convergence of the sum of the reciprocals of the twin primes.
The prime number theorem tells us that P k ∼ k ln ( k ) and therefore P P k ∼ P k ln ( P k ) . For sufficiently large k we have P P k > 2 1 P k ln ( P k ) > 4 1 k ( ln k ) 2 . Now ∑ k = 2 ∞ k ( ln k ) 2 1 converges by the integral test, so that ∑ k = 1 ∞ P P k 1 will converge as well.