Can you prove that $\log \prod_{k\geq 2} \frac{k-\mu(k)}{k+\mu(k)}$ converges?

Question

Since for $|z|<1$ $$\log\frac{1+z}{1-z}=2\sum_{n=0}^{\infty}\frac{z^{2n+1}}{2n+1},$$ and $ |\frac{\mu(k)}{k} |<1$, for $k\geq 2$ where $\mu(k)\in \left\{ -1,0,1\right\} $ is Mobius function then I don't know if this exercise is in the literature $$\log \prod_{k\geq 2} \frac{k-\mu(k)}{k+\mu(k)}=-2 \sum_{k=2}^{\infty} \sum_{n=0}^{\infty}\frac{\mu(k)^{2n+1}}{(2n+1)k^{2n+1}} .$$ Here $0< \frac{k-\mu(k)}{k+\mu(k)}<1$ for $k\geq 1$.

Question. Can you prove that $\log \prod_{k\geq 2} \frac{k-\mu(k)}{k+\mu(k)}$ converges? Thanks in advance.

Answer 1

Since $$\log\left(\frac{1+z}{1-z}\right)=2\sum_{k\geq0}\frac{z^{2k+1}}{2k+1} $$ we have $$S=\log\left(\prod_{n\geq2}\frac{n-\mu\left(n\right)}{n+\mu\left(n\right)}\right)=\sum_{n\geq2}\log\left(\frac{1-\mu\left(n\right)/n}{1+\mu\left(n\right)/n}\right)=-2\sum_{n\geq2}\sum_{k\geq0}\frac{\mu^{2k+1}\left(n\right)}{n^{2k+1}\left(2k+1\right)}$$ $$=-2\sum_{n\geq2}\frac{\mu\left(n\right)}{n}\sum_{k\geq0}\frac{1}{n^{2k}\left(2k+1\right)}.$$ The last sum is easy to calculate, using the geometric series $$\sum_{k\geq0}\frac{1}{n^{2k}\left(2k+1\right)}=\frac{n}{2}\log\left(1+\frac{2}{n-1}\right)$$ and so $$S=-\sum_{n\geq2}\mu\left(n\right)\log\left(1+\frac{2}{n-1}\right). $$ Now recalling that an equivalent form of PNT is $$\sum_{n\geq1}\frac{\mu\left(n\right)}{n}=0\tag{1} $$ we have, using Abel's summation, that $$S=-\sum_{n\geq2}\frac{\mu\left(n\right)}{n}n\log\left(1+\frac{2}{n-1}\right)=2+\int_{2}^{\infty}\left(\sum_{2\leq n\leq t}\frac{\mu\left(n\right)}{n}\right)\left(t\log\left(1+\frac{2}{t-1}\right)\right)^{'}dt $$ and so from $(1)$ we have $$S\sim2-\int_{2}^{\infty}\left(t\log\left(1+\frac{2}{t-1}\right)\right)^{'}dt=2\log\left(3\right) $$ and so the series converges and $2\log\left(3\right)$ is an approximation of that sum.

Answer 2

Thus the full answer from the user in comments is that yes, $$\log \left(\prod_k 1 - \frac{2 \mu(k)}{k} + \mathcal{O}(k^{-2}) \right) = \sum_k \log \left(1 - \frac{2 \mu(k)}{k} + \mathcal{O}(k^{-2})\right)$$ and $$\sum_k \log (1 - \frac{2 \mu(k)}{k} + \mathcal{O}(k^{-2})= \sum_k - \frac{2 \mu(k)}{k} + \mathcal{O}(k^{-2})= \mathcal{O}(1)$$ since $\log(1-x)= x+O(x^2) $ when $x\to 0$, and the last identity by Prime Number Theorem.