Beyond PNT..

lokofer

In fact if PNT says that the series [tex] \sum_{p<x}1 \sim Li(x) [/tex]

My question is if we can't conjecture or prove that:

[tex] \sum_{p<x}p^{q} \sim Li(x^{q+1}) \sim \pi(x^{q+1}) [/tex] q>0

In asymptotic notation.....

matt grime

Of course we can conjecture it, Jose (I presume this a new account for eljose). Have *you* tried to prove it? Does it even seem reasonable? Have you run it through a computer at all? Why do you even think it might be true?

shmoe

See
http://www.physicsforums.com/showthr...ght=conjecture

My suggestion to 'eljose' still applies here "...use partial summation and the prime number theorem including keeping track of error terms."

lokofer

The main key is that according to the manual..."Mathematical Handbook of Formulas and Tables"..the integral:

[tex] \int_{2}^{x} dt \frac{t^n }{log(t)}= A+log(log(x))+\sum_{k>0}(n+1)^{k}\frac{log^{k}}{k. k!} [/tex]

Using the properties of the logarithms you get that the series above is just Li(x^{n+1}) , in fact using "this" conjecture and prime number theorem you get the (known) asymptotic result:

[tex] \sum_{i=1}^{N}p_i \sim (1/2)N^2 log(N) [/tex]

for the case n=-1, you get that the "Harmonic prime series" diverges as log(log(x)) ...although the constant i give is a bit different.