## Beyond PNT..

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

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

$$\sum_{p<x}p^{q} \sim Li(x^{q+1}) \sim \pi(x^{q+1})$$ q>0

In asymptotic notation.....
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 Recognitions: Homework Help Science Advisor 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?
 Recognitions: Homework Help Science Advisor 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."

## Beyond PNT..

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

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

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:

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

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.