Using Such that with summation

Question

$$ - x \big( \ln f ( x ) \big) ' = \sum _ { k \ge 1 } \frac { k m _ k x ^ k } { 1 - m _ k x ^ k } = \sum _ { N \ge 1 } x ^ N \sum _ { s | N } m _ { N /s } ^ s \frac N s \text . $$ I want to ask If someone can explain me this equation. I have been trying for quite some time. The right most side of this equation is something I am not being able to understand how we derived that from center part. Other than that, do we have a divide or a such that sign with summation ($s|N$). $$ f ( x ) = \prod _ { k \ge 1 } \left( 1 - m _ k x ^ k \right) \text , $$ The second part of equation we got by taking the logarithmic derivative of above equation.

Answer 1

From the sum of geometric series formula

$$\frac x{1-x}=x+x^2+x^3+\cdots=\sum_{n\ge1}x^n$$

We have

$$\sum _ { k \ge 1 } \frac { k m _ k x ^ k } { 1 - m _ k x ^ k } = \sum _ { k \ge 1 } \sum_{n\ge1}k (m _ k x ^ {k})^n=\sum _ { k \ge 1 } \sum_{n\ge1}k m _ k^n x ^ {nk}$$

Now this form can be "simplified" to a power series if we extract the coefficient for $x^N$ for every $N$. But the exponent of $x$ is $nk$, so we need to consider factors $s$ of $N$.

Writing $n = s$ and $k = \dfrac N s$ we have:

$$[x^N]\sum _ { k \ge 1 } \sum_{n\ge1}k m _ k^s x ^ {nk}=\sum_{s \mid N} \frac Ns m_{N/s}^s=\sum_{s \mid N}m_{N/s}^s \frac Ns$$

where $[x^N]P(x)$ denotes the coefficient of $x^N$ in $P(x)$.

---

It is better to illustrate the above with an example: $N=6$.

We need to find the set of $(n,k)$ such that $nk=6$. We have $(n,k) = (1,6),(2,3),(3,2),(6,1)$.

Hence the coefficient of $x^6$ in the series is:

$$6m_6^1+3m_3^2+2m_2^3+1m_1^6 = \sum_{s\mid 6}\frac6sm_{6/s}^s$$

---

Finally:

$$\sum _ { k \ge 1 } \sum_{n\ge1}k m _ k^s x ^ {nk} = \sum_{N\ge0}x^N\left([x^N]\sum _ { k \ge 1 } \sum_{n\ge1}k m _ k^s x ^ {nk}\right)=\sum_{N\ge0}x^N\sum_{s \mid N}m_{N/s}^s \frac Ns$$