Let $a_n=\frac{\mu(n)}{n^2}$, where $\mu(n)$ is the Möbius function. Since $\sum \left| a_n \right| $ is convergent by the comparison test, then a proposition from analysis ensures that $$\mathcal{S}=\sum_{n=1}^{\infty} \left| \log \left( 1+\frac{\mu(n)}{n^2} \right) \right| $$ is also convergent.

Question. Can you give a good approximation or get in a closed-form $S$? If you use computational methods (this is with a computer), please explain us how you are ensuring a good approximation. My goal is learn if my claims are rights and learn or refresh more easy facts from your answer. Thanks in advance.

One has that $\mathcal{S}$ seems that is around $1.256$. I don't know if the following could be useful facts to compute it:

1) Combining the Prime Number Theorem that claims that $\sum_{n=1}^\infty\frac{\mu(n)}{n}$ converges with sum $0$, with the proposition showing that if one has $b_n=\frac{\mu(n)}{n}$ with $\sum b_n^2$ convergent, by the comparison test it is, then using a proposition of analysis we can claim that $\prod_{n=1}^\infty \left( 1+ \mu(n)/n\right) $ is also convergent.

2) $S$ can be written as the logarithm of a quotient of products, from $$\log 2-\sum_{n\in A}\log \left( 1-\frac{1}{n^2} \right) +\sum_{m\in B}\log \left( 1+\frac{1}{m^2} \right),$$ where $A$ is the set of square-free integers $n$ with an odd number of distinct prime factors, and $B$ is the set of square-free integers $m$ with an even number of distinct prime factors.

3) And we know the approximation $\log(1+x) \approx x $, that we can use in each summand, for $n\geq 2,$ of the series defining $\mathcal{S}$, and the value for $1/\zeta(2)$. Is it possible combine these facts in 3)?

  • $\begingroup$ It is not difficult to prove that $S$ is less or equal to $\log(2) + \frac{3}{4} \approx 1.4431$ but I don't know if it a sufficiently good approximation for you. $\endgroup$ – Marco Cantarini May 17 '16 at 9:16

A trivial approximation can be the following. Since $\left|\frac{\mu\left(n\right)}{n^{2}}\right|<1 $ if $n\geq2 $ we have $$\sum_{n\geq1}\left|\log\left(1+\frac{\mu\left(n\right)}{n^{2}}\right)\right|=\log\left(2\right)+\sum_{n\geq2}\left|\sum_{m\geq1}\frac{\left(-1\right)^{m+1}\mu^{m}\left(n\right)}{n^{2m}}\right|\leq\log\left(2\right)+\sum_{n\geq2}\sum_{m\geq1}\frac{1}{n^{2m}} $$ $$=\log\left(2\right)+\sum_{n\geq2}\frac{1}{n^{2}-1}=\log\left(2\right)+\frac{3}{4}\approx1.4431 $$ since the last series telescopes. I don't know if it is a good approximation for you. If it is not please tell me and I will delete the answer.

  • $\begingroup$ Is a good answer, since you add nice and simple computations that I can learn. I hope one week accepting your answer. Very thanks much. You answer improve much mine, and should be a miracle that someone could compute in a closed form all impossible series that I am asking in this site! Very than much. $\endgroup$ – user243301 May 17 '16 at 10:23

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