5
$\begingroup$

It is well known that, where $p_k$ is the $k$th prime number (this is $2 = p_1 < p_2 < p_3 < \cdots$), the following

Proposition. The series of reciprocals of primes $$\sum_{k=1}^\infty \frac{1}{p_k}$$ diverges.

Too is known the so called harmonic-arithmetic mean inequality

Proposition. For any positive real numbers $a_1,a_2,\ldots,a_N$, we have $$\left(a_1+a_2+\cdots +a_N\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_N}\right)\geq N^2.$$

Putting $a_k=p_k$ for $1\leq k\leq N$ we obtain $$0<\frac{N^2}{\sum_{k=1}^N p_k}\leq \sum_{k=1}^N \frac{1}{p_k}.$$

Thus $$\lim_{N\to \infty}\frac{N^2}{\sum_{k=1}^N p_k}\leq \infty.$$

My

Question. a) Compute previous limit. b) Compute $\lim_{N\to \infty} \frac{N^\alpha\cdot (\log N)^\beta}{\sum_{k=1}^N p_k}$, where I repeat another time that $p_k$ is the kth prime number and $\alpha,\beta$ real parameters.

Thanks in advance, my only goal is learn in this site Math Stack Exchange from yours answers.

References:

If you want read it, you could find references of proofs of divergence of reciprocals of primes and a proof of the harmonic-arithmetic mean inequality, via this web site or Wikipedia, for example.

$\endgroup$
6
  • $\begingroup$ Do you know that $p_k \sim k\log k$? $\endgroup$
    – Daniel Fischer
    Aug 5, 2015 at 17:59
  • $\begingroup$ Yes, then I understand that it is easy compute partial sums of $\sum k\log k$. $\endgroup$
    – user243301
    Aug 5, 2015 at 18:01
  • $\begingroup$ It's easy to find the asymptotic behaviour. Computing the partial sums exactly is not so easy. $\endgroup$
    – Daniel Fischer
    Aug 5, 2015 at 18:02
  • $\begingroup$ Thanks, Fischer today I shoud try bound these partial sums, or wait that someone compute the asymptotic behaviour with a trick. $\endgroup$
    – user243301
    Aug 5, 2015 at 18:04
  • 1
    $\begingroup$ If you use Dusart's bounds cf. e.g. here, you can even get more than just the leading term of the asymptotics. $\endgroup$
    – Daniel Fischer
    Aug 5, 2015 at 18:09

1 Answer 1

Reset to default
2
$\begingroup$

The sum of the first $N$ primes is asymptotically equal to $\frac12 N^2 \log N$.

Your limit in a) is thus $0$.

For b) it is $0$ for $\alpha < 2$ and any $\beta$ and $\infty$ for $\alpha >2$ and any $\beta$.

If $\alpha= 2$ then it is $0$ for $\beta < 1$, $\infty$ for $\beta >1$ and $2$ for $\beta =1$.

$\endgroup$
1
  • 1
    $\begingroup$ Thanks @quid, I see that I should have search the asymptotic of the sum of first primes​​. Thank you very much for your answer,I vote up your answer now, and after I 'll try to check the cases in b). $\endgroup$
    – user243301
    Aug 5, 2015 at 18:15

You must log in to answer this question.