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How can I evaluate

$$\lim_{k \to \infty} \frac{1}{\log k} ( \sum^k_{p \in primes} \frac{1}{\sqrt{p}} -\sum^k_n \frac{1}{\sqrt{n} \log n})$$

?

My attempt: By the Prime Number Theorem, we can approximate the asymptotic behaviour of the series over prime numbers to $$\frac{\sqrt{k}}{\sqrt{\log k}}$$ Unfortunately, as we are facing a subtraction instead of a division, this seems not useful at all.

Edit 1: This is what I tried: Let's analyse the two sums in the numerator. First, we have

$$\lim_{k \to \infty} \sum^k_n \frac{1}{\sqrt{n} \log n} - li(\sqrt{k}) = Q$$

For a fixed constant $Q$ and $li(x)$ the logarithmic integral.

To work with the other sum, let's define $a_n=\frac{1}{n}$ for $n$ a prime number and $0$ otherwise . Then, by applying Abel's Summation we have that

$$\sum^k_{p \in primes} \frac{\sqrt{p}}{p} = \sqrt{k} \sum_{p \le k} \frac{1}{p} - \frac{1}{2} \int_2^k \sum_{p \le y} \frac{1}{p} \frac{dy}{\sqrt{y}} + Q'$$

For a constant Q'. By applying that $$\lim_{k \to \infty}\sum^k_{p \in primes} \frac{1}{p} - \log \log k = M$$ for a constant $M$, we get:

$$\lim_{k \to \infty} \sum^k_{p \in primes} \frac{\sqrt{p}}{p} - ( \sqrt{k} \log \log k - \sqrt{k} \log \log k + li(\sqrt{k}) ) = Q''$$

For a constant $Q''$. For last, substituting in the original problem, we get that the numerator of our fraction converges to a constant.

user3141592
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    I'm affraid it depends on the Riemann hypothesis. Assume the RH, use $|\pi(x)-li(x)| < C x^{1/2} \ln x$ and the summation by parts – reuns Mar 22 '17 at 10:13
  • @user1952009 Then, would ot only converge if the RH holded? Isn't there any other way to prove its convergence or bounding the function without assuming the hypothesos? – user3141592 Mar 22 '17 at 18:26
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    If the RH is not true then it diverges – reuns Mar 22 '17 at 18:29
  • @user1952009 Why? How can you implement $\pi(x)-li(x)$ using summation by parts? – user3141592 Mar 22 '17 at 18:51
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    $Li(x) = \int_2^x \frac{dt}{\log t} \approx \sum_{n=2}^x \frac{1}{\ln n}$. As $\pi(x) = \sum_{p < x} 1$ it appears using the summation by parts – reuns Mar 22 '17 at 18:54
  • @user1952009 After more research, I found that $$\lim_{k \to \infty} \sum_p^k \frac{1}{\sqrt{p}} \sim li(\sqrt{k})$$. As $$\lim_{k \to \infty} \sum_n^k \frac{1}{\sqrt{n}\log(n)} \sim li(\sqrt{k})$$, wouldn't the numerator of the original limit directly tend to 0? – user3141592 Mar 23 '17 at 19:15
  • No, I think it diverges in any case, and if the RH is not true I'm sure it does. – reuns Mar 23 '17 at 20:29

1 Answers1

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  • Let $L(k) = \sum_{n=2}^k \frac{1}{\ln n}$. Summing by parts : $$\sum_{p \le k} p^{-1/2} = k^{-1/2} \pi(k)+ \sum_{n =1}^{k-1} \pi(n) (n^{-1/2}-(n+1)^{-1/2})$$ $$\sum_{n=2}^k \frac{n^{-1/2}}{\ln n}= k^{-1/2} L(k)+\sum_{n=2}^{k-1} L(n) (n^{-1/2}-(n+1)^{-1/2})$$

  • $n^{-1/2}-(n+1)^{-1/2} = \int_n^{n+1} \frac12 x^{-3/2}dx=\frac{n^{-3/2}}2+O(n^{5/2})$

  • Assume the Riemann hypothesis then $\pi(k) =L(k)+O(k^{1/2}\ln k)$ so that $$\sum_{p \le k} p^{-1/2}-\sum_{n=2}^k \frac{n^{-1/2}}{\ln n} = O(\ln k)+\sum_{n=2}^{k-1} O(n^{1/2}\ln n) (\frac{n^{-3/2}}2+O(n^{5/2})) = O(\ln^2 k)$$

    And hence $$\frac{1}{\color{pink}{\log^2 k}} ( \sum^k_{p \in primes} \frac{1}{\sqrt{p}} -\sum^k_n \frac{1}{\sqrt{n} \log n}) \quad \color{blue}{\text{is bounded}}\tag{1}$$

  • Assume the RH is not true, then $\pi(k)-L(k) \ne O(k^{1/2}\ln k)$ so that $$\sum_{p \le k} p^{-1/2}-\sum_{n=2}^k \frac{n^{-1/2}}{\ln n} \ne O(\ln^2 k) \quad \text{and} \ \ (1) \ \color{red}{\text{is unbounded}}$$

reuns
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  • Sorry, but in (1) you say that the denominator of the fraction is $\frac{1}{\log^2 k}, but the original problem is without that square there. – user3141592 Mar 23 '17 at 20:56
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    @user3141592 It is not a mistake : $\log^2 k$ appears from $\sum_{n < k} \frac{\ln n}{n}$ – reuns Mar 23 '17 at 21:41
  • Thank you for your answer. In spite that it seems correct, I did not "check" it yet because it seems not compatible with my attempt to solve the limit, and I would like to know where my mistake is. I have dediced to update my main question to clarify my intentions. Thanks – user3141592 Mar 24 '17 at 16:27
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    @user3141592 I don't know what you are trying to do. $\sum_{p < k} \frac{1}{p} = \ln \ln k + C + O(k^{-1/2}\ln k)$ again if the RH is true. That's a very difficult subject. – reuns Mar 24 '17 at 16:55
  • Sorry, but I do not see your point. Aren't we discussing about Merten's second theorem? If that k tends to Infinity, the error term disappears – user3141592 Mar 24 '17 at 17:06
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    @user3141592 $\sqrt{k} \sum_{p \le k} \frac{1}{p} =\sqrt{k} \ln \ln k + C\sqrt{k} + O(\ln k)$ iff the RH is true – reuns Mar 24 '17 at 17:17
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    Otherwise $\sqrt{k} \sum_{p \le k} \frac{1}{p} =\sqrt{k} \ln \ln k + C\sqrt{k} + O(k^{\sigma-1/2}\ln k)$ iff $\zeta(s)$ has no zeros on $Re(s) > \sigma$ – reuns Mar 24 '17 at 17:19
  • Ok. I understand what you say, but I do not understand why you say it. My aim there is to prove my limit avoiding RH. Imagine that the error term in Merten's Theorem were $O(k^{-1/100}\ln k)$ (silly example assuming RH were not true). Why wouldn't then my limit work? $\lim_{k \to \infty} k^{-1/100} \ln k =0$ (So sorry for being that stubborn, but I can't get to understand your point) – user3141592 Mar 24 '17 at 19:06
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    @user3141592 Read again what I wrote, correct your mistakes, you'll see why. And no you didn't explain what you want that isn't in my answer (the asymptotics of those complicated things, depending on the RH) – reuns Mar 24 '17 at 19:20