[seqfan] Re: Asymptotics of sum of the first n primes with a remainder term

[Vladimir Shevelev]

In the same way one can prove a generalization for  sum of the first n  k-th powers of primes: 
Sum_{i=1..n} prime(i)^k = 1/(k+1)*n^(k+1)*log(n)^k + O(n^(k+1)*log(n)^(k-1)*log(log(n))).  Except for A007504, I did contributions to A024450, A098999, A122102 and A122103.

Regards,
Vladimir


----- Original Message -----
From: Vladimir Shevelev <shevelev at bgu.ac.il>
Date: Thursday, August 1, 2013 2:57
Subject: [seqfan] Asymptotics of sum of the first n primes with a remainder term
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

> I am writing about A007504. Bach&Shallit proved that a(n) ~ n^2 
> * log(n) / 2. It is not difficult also to prove a more precise 
> formula (with a remainder term). Using the best known Rosser's 
> result for n-th prime: p_n>=n*log(n), we have 
> a(n)=sum[i=1,...,n}p_i>=sum{i=2,...,n}i*log(i)>=int_2^n (t-
> 1)*log(t-1)dt=(n-1)^2log(n-1)/2-(n-1)^2/4+1/4.
> On the other hand, it is known (cf. P.Dusart,arxiv:1002.0442, 
> formulas (4.1),(4.2)) that for n>=4, 
> p_n<=n*(log(n)+log(log(n))+1) and 
> a(n)=sum[i=1,...,n}p_i<=c+sum{i=1,...,n}(i*(log(i)+log(log(i))+1)<=c+(log(n)+log(log(n))+1)n*(n+1)/2 (with constant c). Thus a(n)=n^2 * log(n) / 2 + O(n^2*log(log(n))). 
> 
> Regards,
> Vladimir
> 
>  Shevelev Vladimir‎
> 
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 Shevelev Vladimir‎