Partial sum of the reciprocal of the primes squared

Question

It is known that the sum of the reciprocal of the primes squared converge to $0.4522\cdots$. Here my question is: is

$$\sum_{\genfrac{}{}{0pt}{1}{p>N}{p\text{ is prime}}}\frac1{p^2}<\frac1N$$

hold for every positive integers $N$? I will be appreciated for any useful comments.

Answer 1

$$\sum_{\genfrac{}{}{0pt}{1}{p>N}{p\text{ is prime}}}\frac1{p^2} <\sum_{k>N} \frac1{k(k-1)}= \sum_{k>N} \Bigl(\frac1{ k-1 }-\frac1k\Bigr)=\frac1N \,.$$