"Punkish301" wrote:

> asymptotic formula for

> sum p_n<=x 1/(p_n - p(n-1))

At large x, the probability of primality is asymptotic to

1/log(x) and hence the average gap between primes is log(x).

The integral of 1/log(x)^2 diverges as x/log(x)^2.

However, your sum favours small gaps. So let us consider the

role of the twin primes. They have a conjectural density of

order 1/log(x)^2. Hence their contribution to the sum is, by

itself, of order x/log(x^2). So one might expect that the

full sum diverges faster than x/log(x)^2.

> I've tested this up to 500,000.

We may easily go further.In a few minutes,

I obtained results up to x = 10^10.

For prime p > 2, let g(p) = p - q be the gap between p and

the previous prime, q. Then consider this sum over prime p:

S(x) = sum(2 < p < x, 1/g(p))

Here is a table of [x, S(x)*log(x)^2/x]:

[1000, 1.932409314]

[10000, 2.104903983]

[100000, 2.175330374]

[1000000, 2.252107255]

[10000000, 2.335029340]

[100000000, 2.402612252]

[1000000000, 2.468588116]

[10000000000, 2.530925912]

So it is likely that S(x) increases faster than x/log(x)^2.

David