What is the value of $\sum_{p\le x} 1/p^2$?

Question

My question is that what is the value of $$\sum_{p\le x} \frac{1}{p^2}?$$ More generally, what is the value of $$\sum_{p\le x} \frac{1}{p^n}?$$ How can we find it?

For $\sum_{p\le x} 1/p$ the idea is to use the summation $\sum_{p\le x}\ln p/p$ and Abel's summation formula. Actually, can we find any expression for something like my question? When I use Abel's summation formula for $\sum_{p≤x}1/p$ and $\sum_{p≤x}\ln(p)/p$ some hard integrals appear.

Would you please help me? Thank you a lot!

Maybe this could be helpful http://math.stackexchange.com/questions/185080/partial-summation-of-a-harmonic-prime-square-series-prime-zeta-functions — kingW3, Jan 16 '15 at 21:02
See http://math.stackexchange.com/questions/99007/rate-of-convergence-of-series-of-squared-prime-reciprocals. — Dietrich Burde, Jan 16 '15 at 21:41
This sum converges to a finite number, but there is no nice closed form for it. Its decimal expansion is given here: https://oeis.org/A085548 — Peter Humphries, Jan 17 '15 at 15:55
With $p$ you're talking about prime numbers, right? That's my first instinct, plus there's the (prime-numbers) tag, but still, it wouldn't hurt to say so more explicitly. — Robert Soupe, Jan 21 '15 at 02:55
I contend this isn't a duplicate, because this question asks about $\sum 1/p^n$, not just $\sum 1/p^2$. — apnorton, Jan 21 '15 at 05:35
See also https://mathoverflow.net/questions/53443/sum-of-the-reciprocal-of-the-primes-squared — Gerry Myerson, Apr 28 '17 at 09:17

score 3 · Answer 1 · answered Feb 14 '15 at 09:08

You probably wont be able to find a closed form representation for these sums in terms of functions you would consider simple or elementary. However if $\Re(n)>1$ and we define the following constant:

$$C_n=\sum_{k=2}^\infty \frac{\mu(k)\ln(\zeta(nk))}{k}$$

Then you can rewrite your sum as:

$$\sum_{p\leq x}\frac{1}{p^n}=\sum_{p}\frac{1}{p^n}-\sum_{p>x}\frac{1}{p^n}=C_n+\frac{\pi(x)}{x^n}-n\int_{x}^\infty \frac{\pi(t)}{t^{n+1}} dt$$

Which if we assume the truth of RH would give us that:

$$\sum_{p\leq x}\frac{1}{p^n}=C_n+\frac{\text{Li}(x)}{x^n}-n\int_{x}^\infty\frac{\text{Li}(t)}{t^{n+1}}dt+O(\frac{\ln(x)}{x^{n-1/2}})$$

What is the value of $\sum_{p\le x} 1/p^2$?

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