## Does sum log(p)/(p(p-1)) converge?

Expand Messages
• 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 =
Message 1 of 3 , Feb 9, 2005
• 0 Attachment
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
Message 2 of 3 , Feb 9, 2005
• 0 Attachment
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]
• ... This formula is wrong. Please replace it by B3 = gamma + sum_{p prime} log(p)/(p*(p-1)) where B3 = log x - sum_{p prime
Message 3 of 3 , Feb 9, 2005
• 0 Attachment
On Wednesday 09 February 2005 14:18, you wrote:
> 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)).

This formula is wrong. Please replace it by

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]
Your message has been successfully submitted and would be delivered to recipients shortly.