What value does the sum of the square of the reciprocals of the prime counting function converge to?

Question

Using my only mathematical tool Wolfram Alpha, I noticed that $\sum_{n=2}^\infty \frac{1}{\pi(x)}$ seems to diverge. Naturally, the entered the following sum and, just like the zeta function, saw that it converges when each term is squared! The partial sums from WA suggest that it converges to a number greater than $e$ and would like to know if a closed form is known.

$$\sum_{x=2}^\infty \frac{1}{{\pi(x)}^2} = ?$$

Answer 1

I don't know if there is a more elementary way to do it but I am going to risk total overkill by using (part of) the Prime Number Theorem. There is a constant $c>0$, in fact any $c<1$ will do, such that $$\pi(x)>\frac{cx}{\log x}$$ for all sufficiently large $x$. Therefore $$\frac{1}{\pi(n)^2}<\frac{1}{c^2}\frac{(\log n)^2}{n^2}\ ,$$ for all large $n$, and the series converges by comparison with $$\sum_{n=2}^\infty\frac{1}{n^{2-\varepsilon}}\ .$$

In fact a similar argument will show that for any $a>1$, the series $$\sum_{n=2}^\infty\frac{1}{\pi(n)^a}$$ converges. For $a=1$, the first series you looked at, it does indeed diverge because the terms include all of the numbers $$\frac{1}{1}\,,\ \frac{1}{2}\,,\ \frac{1}{3}\,,\ldots\ ,$$ with many of them repeated.

Answer 2

I doubt that there's a closed form, but you might notice that $$\sum_{n=2}^\infty \dfrac{1}{\pi(n)^2} = -2 + \sum_{j=2}^\infty \left( \dfrac{1}{(j-1)^2} - \dfrac{1}{j^2} \right) p_j$$ where $p_j$ is the $j$'th prime.