Argument
Let, $f(x) = \sum_{r=1}^\infty \mu(r) x^r$ where $\mu(r)$ is the mobius function. Hence,
$$ f(x) + f(x^2) + \dots = x$$
Now, let $x \to x^2$:
$$ 0+f(x^2) + 0+\dots = x^2$$
Similarly $x \to x^3$ :
$$ 0 + 0 + f(x^3) + \dots = x^3 $$
$$\vdots$$
$$ \underbrace{0 + 0}_{n-1 \text{ times}} + f(x^n) + \dots = x^n $$
Adding the above (vertically) we get defining $\kappa(n)$ as number of factors of $n$:
$$ \kappa(1) f(x) + \kappa(2) f(x^2) + \kappa(3) f(x^3) + \dots = \frac{x}{1-x}$$
Defining $g(x) = f(x) \ln x$
$$ \kappa(1) \frac{g(x)}{\ln x} + \kappa(2) \frac{g(x^2)}{2 \ln x} + \kappa(3) \frac{g(x^3)}{3 \ln x} + \dots = \frac{x}{1-x}$$
Multiplying $\ln(x)$ both sides:
$$ \kappa(1) \frac{g(x)}{1} + \kappa(2) \frac{g(x^2)}{2 } + \kappa(3) \frac{g(x^3)}{3 } + \dots = \frac{x \ln x}{1-x}$$
Let, $x \to 1 - \epsilon$ and keeping $\epsilon$ in first order and below terms:
$$ \kappa(1) \frac{g(x)}{1} + \kappa(2) \frac{g(x^2)}{2 } + \kappa(3) \frac{g(x^3)}{3 } + \dots = \frac{x \ln x}{1-x}$$
Taking limit $x \to 1 $ both sides:
$$ g(1) (\kappa(1) + \frac{\kappa(2)}{2} + \frac{\kappa(3)}{3}) + \dots) = -1 $$
But due to result $1$ we know that the series in result $1$ that in the limit $n \to \infty$ the series is undefined! Now, $g(1)$ is obviously outside the radius of convergence. Is this evidence there may exist any analytic form of $g(x)$ which can by analytically continued outside the radius of convergence to satisfy the equation below?
$$ g(1) (\kappa(1) + \frac{\kappa(2)}{2} + \frac{\kappa(3)}{3}) + \dots) = -1 $$
Result $1$
Using asymptotic formula for a harmonic series:
$$1+\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} \sim \gamma + \ln n$$
where $\gamma$ is euler's constant. Replacing with $n \to n/2 +O(1)$ and multiplying by $\frac{1}{2}$:
$$ 0+ \frac{1}{2} +\dots +\frac{1}{n} \sim \frac{\gamma}{2} + \frac{\ln( n/2 +O(1))}{2}$$
Replacing with $n \to n/3 +O(1)$ and multiplying by $\frac{1}{3}$:
$$ 0+ 0 + \frac{1}{3} +\dots +\frac{1}{n} \sim \frac{\gamma}{3} + \frac{\ln( n/3 +O(1))}{3}$$
$$\vdots$$
$$ \underbrace{0+ 0 }_{n-1 \text{ times}}+ \dots + \frac{1}{n} \sim \frac{\gamma}{n} + \frac{\ln( 1 +O(1))}{n}$$
Adding the above (vertically) we get and using $1+\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} \sim \gamma + \ln n$:
$$ \kappa(1) + \frac{\kappa(2)}{2} + \frac{\kappa(3)}{3} + \dots + \frac{\kappa(n)}{n} \sim \gamma (\ln n +\gamma) + \sum_{r=2}^n \ln(\frac{n}{r} (1+ \frac{r}{n}(O(1)) $$
Simplifying further:
$$ \kappa(1) + \frac{\kappa(2)}{2} + \frac{\kappa(3)}{3} + \dots + \frac{\kappa(n)}{n} \sim \gamma (\ln n +\gamma) + n \ln(n) - \ln(n!) +\sum_r^n \ln (1+ \frac{r}{n}(O(1)) $$