One of three of a kind - Part (2)

So, this is our problem -

$\displaystyle \sum_{k=1}^\infty \frac{\mu(k)}k \ln(1 - x^k)$

$\displaystyle - \sum_{k=1}^{\infty}{\frac{\mu(k)}{k} \sum_{m=1}^{\infty}{\frac{x^{km}}{m}}}$

We will just use Dirichlet Convolution or just transforming the sum to the variable $km = n$ ,

$\displaystyle - \sum_{n=1}^{\infty}{\sum_{k|n}{\frac{\mu(k)}{k}} \frac{x^n}{n}}$

Now, we know that $\mu(n)$ is just the Mobius Function i.e. sum of primitive nth roots of unity. It is just

$= 1 \text{if n is a square-free positive integer with even prime factors}$

$= -1 \text{if n is a square-free positive integer with odd prime factors}$

$= 0 \text{if it is a square number}$

Hence, the sum we have used is just

$= 1 \text{if n=1}$

$= 0 \text{if n>1}$

As a result,

$\displaystyle - \sum_{n=1}^{\infty}{\sum_{k|n}{\frac{\mu(k)}{k}} \frac{x^n}{n}} = - \frac{x^1}{1} + 0 = -x$