Is there a closed form for $$\sum_{n=2}^{\infty} \frac{1}{\pi (n)^2}$$ (where $\pi (x)$ is the number of primes less than or equal to x)?
It's obvious that $$\sum_{n=2}^{\infty} \frac{1}{\pi (n)^s}$$ converges for $s>1$ by observing the inequality $\pi(x)>\frac{x}{\ln(x)}>x^k$ for $k<1$ and $x$ greater than some finite positive integer. If any extra information can be given about the generalization $\sum_{n=2}^{\infty} \frac{1}{\pi (n)^s}$, that would be great too.
EDIT:
So far, I've discovered the following
Define $$f(s)=\sum_{n=2}^{\infty} \frac{1}{\pi (n)^s}$$
Then $f(s)$ is equal to the Dirichlet series $$\sum_{n=1}^{\infty} \frac{g_n}{n^s},$$ where $g_n$ is the nth prime gap.
Then by using Abel's summation formula, $f(s)$ is equal to the improper integral $$s \int_1^{\infty} \frac{p_{\lfloor u+1 \rfloor}-2}{u^{1+s}} du,$$ where $p_n$ is the nth prime number.
Hope that helps