# Asymptotic behaviour of $\sum_{p\leq x} \frac{1}{p^2}$

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:

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

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

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The sum is bounded, so asymptotically it is constant. –  Alex Becker Jul 9 '12 at 3:02
@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. –  robjohn Jul 11 '12 at 19:44

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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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?! –  MathematicalPhysicist Jul 9 '12 at 3:42
@MathematicalPhysicist I don't understand your comment. Your sum $\sum_{p = 1}^{\infty} \frac{1}{p^2}$ converges. Hence, your sum $\sum_{p \leq x} \frac{1}{p^2} = \mathcal{O}(1)$. –  user17762 Jul 9 '12 at 3:55
OK, thanks. sorry for the misunderstanding from my behalf. –  MathematicalPhysicist Jul 9 '12 at 4:50

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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