- According to PARI/GP,

sum_{p prime} log(p)/(p*(p-1))

appears to converge.

? maxp = 100000; s = 0.; forprime(p=2,maxp,s+=log(p)/(p*(p-1))); s

%1 = 0.7553566278090918349304694602

? maxp = 1000000; s = 0.; forprime(p=2,maxp,s+=log(p)/(p*(p-1))); s

%2 = 0.7553656108009827466628435907

? maxp = 10000000; s = 0.; forprime(p=2,maxp,s+=log(p)/(p*(p-1))); s

%3 = 0.7553665108289091347321424396

? maxp = 100000000; s = 0.; forprime(p=2,maxp,s+=log(p)/(p*(p-1))); s

%4 = 0.7553666008318318260357569144

Any thoughts?

I might share my interest on this sum with the list, if anyone cares. Just a

hint: 1/p^2 + 1/p^3 + 1/p^4 + ... = 1/(p(p-1)).

Décio

[Non-text portions of this message have been removed] - Folks, don't waste your time on this one. David Broadhurst pointed me again to

his method to compute sum_{p prime} log(p)/p, and one of the intermediate

steps in his derivation is

sum_{p prime} log(p)/p = gamma + sum_{p prime} log(p)/(p*(p-1)).

Thus the sum converges, and due to David's work it is known to 10,000 digits

of accuracy.

I'm sorry for wasting your time and for being so stupid as to forget David's

wonderful method after just a week.

Décio

On Wednesday 09 February 2005 13:42, you wrote:

> According to PARI/GP,

>

> sum_{p prime} log(p)/(p*(p-1))

>

> appears to converge.

>

> ? maxp = 100000; s = 0.; forprime(p=2,maxp,s+=log(p)/(p*(p-1))); s

> %1 = 0.7553566278090918349304694602

> ? maxp = 1000000; s = 0.; forprime(p=2,maxp,s+=log(p)/(p*(p-1))); s

> %2 = 0.7553656108009827466628435907

> ? maxp = 10000000; s = 0.; forprime(p=2,maxp,s+=log(p)/(p*(p-1))); s

> %3 = 0.7553665108289091347321424396

> ? maxp = 100000000; s = 0.; forprime(p=2,maxp,s+=log(p)/(p*(p-1))); s

> %4 = 0.7553666008318318260357569144

>

> Any thoughts?

>

> I might share my interest on this sum with the list, if anyone cares. Just

> a hint: 1/p^2 + 1/p^3 + 1/p^4 + ... = 1/(p(p-1)).

>

> Décio

[Non-text portions of this message have been removed] - On Wednesday 09 February 2005 14:18, you wrote:
> Folks, don't waste your time on this one. David Broadhurst pointed me again

This formula is wrong. Please replace it by

> to his method to compute sum_{p prime} log(p)/p, and one of the

> intermediate steps in his derivation is

>

> sum_{p prime} log(p)/p = gamma + sum_{p prime} log(p)/(p*(p-1)).

B3 = gamma + sum_{p prime} log(p)/(p*(p-1))

where B3 = log x - sum_{p prime <= x} log(p)/p as x goes to infinity.

Décio

[Non-text portions of this message have been removed]