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:

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

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

Thanks in advance.

Solutions Collecting From Web of "Asymptotic behaviour of $\sum_{p\leq x} \frac{1}{p^2}$"

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.

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 $m$

Summing by parts and using $(1)$ with $m=2$, we get
Using $(2)$, we get
\sum_{p\ge n}\frac{1}{p^2}