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I will not be surprised if this question seems trivial in MO but I asked it first in MathSE and I did not get an answer. Here is the question:
I would like to know if there is a good estimate for the sum which concerns all primes not exceeding $x$:
$$\sum_{p\leq x}\frac{1}{p^s}$$ with $0 < s < 1$. I have trouble bounding the sum from above.

Thanks in advance!

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    $\begingroup$ This question is not reserch level question. But you can find nice bounds for prime zeta function in this doctoral thesishttp://www.math.sjsu.edu/~goldston/tsang1.pdf $\endgroup$
    – user21574
    Commented Feb 22, 2014 at 15:13
  • $\begingroup$ @HassanJolany I know it is not.As I mention above i hoped that someone here might have the kind intention to help me and answer my question.Thank you for the responce! $\endgroup$
    – Konstantinos Gaitanas
    Commented Feb 22, 2014 at 15:21
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    $\begingroup$ "I will not be surprised if this question seems trivial in MO but i asked it first in MathSE and i did not get an answer." In fact, it has already been asked and answered several times on math.stackexchange. See here for a comprehensive answer which shows that for $k>-1$, $$\sum_{p\leq x}p^{k}=\text{li}\left(x^{k+1}\right)+O\left(x^{k+1}e^{-c\sqrt{\log x}}\right).$$ For this reason I vote to close this question. $\endgroup$
    – Eric Naslund
    Commented Apr 26, 2014 at 20:40

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I will give the following $$\sum_{p\leq x}\frac{1}{p^s} \approx \frac{1}{1-s}\frac{x^{1-s}}{\log x}$$ for $0\leq s < 1$ and $$\sum_{p\leq x}\frac{1}{p} \approx \log \log x$$ for $s = 1$.

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    $\begingroup$ Thank you very much for the responce!What methods did you use?Is it possible to arrive at this estimate without the prime number theorem?Using Chebyshev estimates for example $\endgroup$
    – Konstantinos Gaitanas
    Commented Feb 22, 2014 at 7:59
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This estimate is probably very crude, but here we have that $\sum_{p \leq x} \frac{1}{p^s} \leq \pi(x)\frac{1}{2^s}$.

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    $\begingroup$ For such a crude estimate it is not that bad,but it is really far from truth. $\endgroup$
    – Konstantinos Gaitanas
    Commented Feb 22, 2014 at 9:03

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