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


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.


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.

  • $\begingroup$ Do you know that $p_k \sim k\log k$? $\endgroup$ Commented 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
    Commented Aug 5, 2015 at 18:01
  • $\begingroup$ It's easy to find the asymptotic behaviour. Computing the partial sums exactly is not so easy. $\endgroup$ Commented 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
    Commented 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. $\endgroup$ Commented Aug 5, 2015 at 18:09

1 Answer 1


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 $\beta =1$.

  • 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
    Commented Aug 5, 2015 at 18:15

You must log in to answer this question.