| Andrea Ribuoli on Thu, 07 Jan 2021 21:05:23 +0100 | 
[Date Prev] [Date Next] [Thread Prev] [Thread Next] [Date Index] [Thread Index]
| Curiosity | 
| Good evening,   I have installed PariGP and tried -in sequence- the following calculations (manually extending the number of iterations):             c=0.0; for (i=1,    1000000, c=c+bigomega(i); print(c/i))   3.626619...             c=0.0; for (i=1,   10000000, c=c+bigomega(i); print(c/i))   3.786124...             c=0.0; for (i=1,  100000000, c=c+bigomega(i); print(c/i))   3.923512...             c=0.0; for (i=1, 1000000000, c=c+bigomega(i); print(c/i))   4.044220... Is it possible to use PariGP to guess the limit of (c/i) for i => ∞ ? Regards,   Andrea |