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This is fairly straight forward: $$\sum_{p\space\text{prime}}^x \frac{1}{p_x} \sim \ln(\ln(x))$$

And if $$\sum_{c\space \text{composite}}^x \frac{1}{c_x}\sim f(x)$$

Then what is $f(x)$?

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    You subtract the sum of the reciprocals of the primes from the sum of all reciprocals (minus $1$, since $1$ is neither prime nor composite). What do you get? – Daniel Fischer Jul 19 '15 at 18:16
  • @DanielFischer Oh... Really didn't think of that one. So, it follows: $\ln(x)-\ln(\ln(x))$? –  Jul 19 '15 at 18:21
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    If by $\sim$ you mean asymptotic equality, then just $\ln (x)$ suffices. But we have better expressions for $\sum \frac{1}{p}$ and $\sum \frac{1}{n}$, so we can get a better expression for $\sum \frac{1}{c}$, like $$\ln (x) - \ln (\ln x) + (\gamma - 1 - M) + O((\ln x)^{-1}),$$ where $\gamma$ is the Euler Mascheroni constant, and $M$ the Meissel-Mertens constant. – Daniel Fischer Jul 19 '15 at 18:26

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Since every positive integer is either $1$, a prime, or composite, we have

$$\sum_{\substack{c \leqslant x\\ c\text{ composite}}} \frac{1}{c} = \sum_{2 \leqslant k \leqslant x} \frac{1}{k} - \sum_{\substack{p\leqslant x\\ p \text{ prime}}} \frac{1}{p}.$$

For the first sum we know

$$\sum_{2\leqslant k \leqslant x} \frac{1}{k} = \ln (x) + \gamma - 1 + O(x^{-1})$$

where $\gamma$ is the Euler-Mascheroni constant. For the sum of the reciprocals of the primes, Mertens' second theorem tells us

$$\sum_{p \leqslant x} \frac{1}{p} = \ln (\ln x) + M + O((\ln x)^{-1}),$$

where $M$ is the Meissel-Mertens constant, so that together we find

$$\sum_{\substack{c \leqslant x\\ c\text{ composite}}} \frac{1}{c} = \ln (x) - \ln (\ln x) + (\gamma - 1 - M) + O((\ln x)^{-1}).$$

If $1$ is to be considered composite, the subtraction of $1$ is of course to be omitted.

Daniel Fischer
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