I am trying to find a closed approximation of partial reciprocal sum of squared primes, $ \sum_{p \leq n} \frac{1}{p^2} $.
As far as I know, approximation of $ \sum_{p \leq n} \frac{1}{p} $ is well-known, and it can be approximated to $ \sum_{k=2}^{n} \frac{1}{k \log(k)}$ or $ \int_{2}^{n} \frac{dx}{x \log(x)}$. I also know that $\sum \frac{1}{p} $ diverges and $ \sum \frac{1}{p^2} $ converges.
However, I could not find nor derive an approximation of partial reciprocal sum of squared primes.
Is there anyone who knows what an approximated form of $ \sum_{p \leq n} \frac{1}{p^2} $ is?
Below here is my approach to the approximation, and I think there might be several mistakes.
$$ \begin{aligned} \sum_{p \leq n} \frac{1}{p^2} &= \sum_{k=1}^{n} \frac{\pi(k^2)-\pi(k^2-1)}{k^2} \\ &= \sum_{k=1}^{n} \frac{\pi(k^2)}{k^2} - \sum_{k=0}^{n-1} \frac{\pi(k^2)}{k^2+1} \\ &= \frac{\pi(n^2)}{n^2} + \sum_{k=1}^{n-1} \frac{\pi(k^2)}{k^2(k^2+1)} \\ &= \frac{\pi(n^2)}{n^2} + \sum_{k=1}^{n-1} \frac{\pi(k^2)}{k^3} - \sum_{k=1}^{n-1} \frac{\pi(k^2)}{k^2(k^2+1)} \\ &\approx \frac{1}{\log^2(n)} + \sum_{k=2}^{n-1} \frac{1}{k\cdot \log^2(k)} - \sum_{k=2}^{n-1} \frac{1}{(k^2+1) \cdot \log^2(k)} \\ &\approx \sum_{k=2}^{n-1} \frac{1}{k\cdot \log^2(k)} \\ &\approx \int_{2}^{n} \frac{dx}{x\cdot \log^2(x)} \end{aligned} $$
Here, I guess I made a mistake in the fourth line or later, but I do not know why it is a mistake.
Please let me know which part of my steps should be corrected.
