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As the title suggests, I want to find the asymptotic behaviour of this sum as $x\rightarrow \infty$, I tried by summation by parts but didn't succeed I also tried using the asymptotic behvaiour of the sum

$$\sum_{p\leq x} \frac{1}{p} \sim_{x \to \infty} \log \log x$$

i.e squaring both sides gives me:

$$\sum_{p\leq x} \frac{1}{p^2} + \sum_{\substack{q,p\leq x\\p\neq q}} \frac{1}{pq} \sim_{x \to \infty} \log^2(\log x)$$

But then, how do I estimate the second term in the LHS?

Thanks in advance.

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    $\begingroup$ The sum is bounded, so asymptotically it is constant. $\endgroup$ – Alex Becker Jul 9 '12 at 3:02
  • $\begingroup$ @AlexBecker: I took the approach of looking at the asymptotic behavior of the tail; that is, the difference between the partial sum and the limit. $\endgroup$ – robjohn Jul 11 '12 at 19:44
  • $\begingroup$ See also: math.stackexchange.com/questions/99007/… $\endgroup$ – Martin Sleziak Nov 8 '15 at 6:13
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The prime zeta function $P(s)$, for $\text{Real}(s) > 1$, is defined as $$P(s) = \sum_{\overset{p=1}{p \text{ is prime}}}^{\infty} \dfrac1{p^s}$$ The sum converges for $\text{Real}(s) > 1$, similar to the $\zeta$-function. Your sum is $P(2)$ and is approximately $0.4522474200410654985065\ldots$.

There are no "nice" values for $P(s)$ where $s \in \mathbb{Z}^+ \backslash \{1\}$. A very crude argument why there are no "nice" values for $P(s)$ is due to the fact that the function, $$g(n) = \dfrac{\mathbb{I}_{n \text{ is a prime}}}{n^s}$$ is not a "nice" arithmetic function in the usual sense i.e. it is not even multiplicative for instance.

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  • $\begingroup$ This function should have an asymptotic behaviour of $c \log^2\log x$, cause I have been asked to calculate the 4-th moment of the function $R_X(t)=\sum_{p\leq x} \frac{\sin(t\log p)}{\sqrt{p}}$, and to show that it's asymptotically behaves, for $x\in [T^{\delta},T^{1/2-\delta}]$ where $\delta \in (0,1/4)$ , like $c \log^2\log T$ for $T\rightarrow \infty$. In my calculations I ma led to find the sum $\sum_{p\leq x} \frac{1}{p^2}$ asymptotically. any tips as to how to solve this problem?! $\endgroup$ – MathematicalPhysicist Jul 9 '12 at 3:42
  • $\begingroup$ @MathematicalPhysicist I don't understand your comment. Your sum $\sum_{p = 1}^{\infty} \frac{1}{p^2}$ converges. Hence, your sum $\sum_{p \leq x} \frac{1}{p^2} = \mathcal{O}(1)$. $\endgroup$ – user17762 Jul 9 '12 at 3:55
  • $\begingroup$ OK, thanks. sorry for the misunderstanding from my behalf. $\endgroup$ – MathematicalPhysicist Jul 9 '12 at 4:50
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Using an estimate of the difference between the prime counting function, $\pi(n)$ and the logarithmic integral function, $\mathrm{li}(x)$, we can estimate the tail of the series.

Let $\Delta(n)=\pi(n)-\mathrm{li}(n)$. Without assuming the Riemann Hypothesis, we have that for any $m$ $$ \Delta(n)=O\left(\frac{\raise{2pt}n}{\log(n)^m}\right)\tag{1} $$

Summing by parts and using $(1)$ with $m=2$, we get $$ \begin{align} \sum_{k=n}^\infty\frac{1}{k^2}(\Delta(k)-\Delta(k-1)) &=-\frac{\Delta(n-1)}{n^2}+\sum_{k=n}^\infty\Delta(k)\left(\frac{1}{k^2}-\frac{1}{(k+1)^2}\right)\\ &=O\left(\frac{1}{n\log(n)^2}\right)\tag{2} \end{align} $$ Using $(2)$, we get $$ \begin{align} \sum_{p\ge n}\frac{1}{p^2} &=\sum_{k=n}^\infty\frac{1}{k^2}(\pi(k)-\pi(k-1))\\ &=\sum_{k=n}^\infty\frac{1}{k^2}(\mathrm{li}(k)-\mathrm{li}(k-1))+O\left(\frac{1}{n\log(n)^2}\right)\\ &=\int_n^\infty\frac{\mathrm{d}x}{x^2\log(x)}+O\left(\frac{1}{n\log(n)^2}\right)\\ &=\int_n^\infty\frac{\log(x)+1}{x^2\log(x)^2}\mathrm{d}x+O\left(\frac{1}{n\log(n)^2}\right)\\ &=\frac{1}{n\log(n)}+O\left(\frac{1}{n\log(n)^2}\right)\tag{3} \end{align} $$

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  • $\begingroup$ Do you have expertise in the stuff related to RH? If so, I would be interested if you can shoot down this claimed proof, since there doesn't seem to be any published error (even though it should be easy since it's been so many years). If you ask me to bet, I would of course bet it is wrong, but I am just curious how bad the proof is haha. I myself know nothing about RH... $\endgroup$ – user21820 Jul 2 at 11:28
  • $\begingroup$ @user21820: I have very little expertise related to RH. I will take a look, but I doubt I will be able to find anything if no one else has in almost 8 years. $\endgroup$ – robjohn Jul 2 at 15:17
  • $\begingroup$ I see, thanks! I suspect it's just because it just happened to miss everyone's notice. $\endgroup$ – user21820 Jul 2 at 19:31

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