I just come across this question by trying to analyze the pseudoinverse of some infinite matrix (the matrix T as interpreted in my answer to this MSE-question), where this series occurs from some dotproducts. I think it can also be expressed in terms of the Moebius-function: $$\sum_{k=1}^\infty \left| {\operatorname{moebius}(k) \over k} \right| $$ Remembering, that the sum of the reciprocals of the primes is divergent I think I should assume divergence here as well, but I'm not sure.

  • Q1: Is this sum convergent (and if, what is its value)?

  • Q2: if it is divergent, is there possibly some finite value related, like for instance the Euler/Mascheroni-$\gamma$ for the harmonic series?

  • $\begingroup$ It's divergent because it's at least $\sum_p\frac{1}{p}$. $\endgroup$ – Unit May 8 '15 at 7:56
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    $\begingroup$ Well... prime numbers are certainly square-free, thus $$\sum_{p \; \text{prime}} \frac{1}{p} \leq \sum_{k = 1}^{\infty} \frac{|\mu(k)|}{k}$$... $\endgroup$ – A.P. May 8 '15 at 7:56
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    $\begingroup$ $\displaystyle\sum_{n\text{ squarefree}} \dfrac{1}{n} = \displaystyle\prod_{p\text{ prime}} \left(1+\dfrac{1}{p}\right) $ $\endgroup$ – Darth Geek May 8 '15 at 8:00
  • $\begingroup$ @A.P. : <arrgh> reading your argument makes me feeling nearly stupid (but well, it's ok). Of course! (I asked question Q2 there more out of couriosity.... ) $\endgroup$ – Gottfried Helms May 8 '15 at 8:04
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    $\begingroup$ @Darth: thanks too - that formal re-expression of the sum is a nice reminder to look at - perhaps it helps to understand things better if needed later. $\endgroup$ – Gottfried Helms May 8 '15 at 8:06

The asymtotic growth of the sum of the reciprocals of squarefree numbers $n\le x$ is $\frac{6 e^\gamma}{\pi^2} \log x$, because we have

$$ \prod\limits_{p≤x}\left(1 + \frac{1}{p} \right) = \frac{\prod\limits_{p≤x}\left(1 - \frac{1}{p^2} \right)}{\prod\limits_{p≤x}\left(1 - \frac{1}{p} \right)} \sim \frac{\frac{1}{\zeta(2)}}{\frac{1}{e^\gamma \log x}} = \frac{6 e^\gamma}{\pi^2} \log x. $$ In particular we have $$ \sum_{n {\rm \; squarefree}}\frac{1}{n}=\prod_{p}\left(1 + \frac{1}{p} \right) =\infty. $$

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