Understanding a proof of an Identity involving the Möbius Function

Question

Lemma: For each integer $N\geq 1$, we have

$$\sum_{(k,N)=1}\frac{t^k}{k}=-\sum_{d|N}\frac{\mu(d)}{d}log(1-t^d)$$

Proof. Let $c_{k}=1$ if $(k,N)=1$ and $c_{k}=0$ otherwise. By basic propierties of the Möbius function we have $c_{k}=\sum_{d|(k,N)}\mu(d)$. Hence

$$\sum_{(k,N)=1}\frac{t^k}{k}=\sum_{k=1}^{\infty}\frac{t^k}{k}$$

$$=\sum_{k=1}^{\infty}\left(\sum_{d|(k,N)}\mu(d)\right)\frac{t^k}{k}$$ $$=\sum_{d|N}\mu(d)\sum_{k\geq 1,\ d|k} \frac{t^k}{k} \ (\text{How can I do this?})$$ $$=\sum_{d|N}\mu(d)\sum_{k'=1}^{\infty}\frac{t^{dk'}}{dk'}$$ $$=\sum_{d|N}\frac{\mu(d)}{d}\sum_{k'=1}^{\infty}\frac{(t^{d})^k}{k}$$ $$=-\sum_{d|N}\frac{\mu(d)}{d}log(1-t^d)$$ This complete the proof of Lemma.

My question is How can I write

$$\sum_{k=1}^{\infty}\left(\sum_{d|(k,N)}\mu(d)\right)\frac{t^k}{k}$$

in the form $$=\sum_{d|N}\mu(d)\sum_{k\geq 1,\ d|k} \frac{t^k}{k}$$?

Answer 1

$d$ devides $k$ and $d$ divides $n$. Furthermore $\mu(d)$ does not depend on $k$, so we can just sum over $d \vert N$. For $t^k /k$ we have to sum from $k=1$ to $k = \infty$ but only for those $k$ that are divisible by $d$.

Answer 2

$$\sum_{k=1}^{\infty}\left(\sum_{d|(k,N)}\mu(d)\right)\frac{t^k}{k}$$ means the summ when your idices run over the set $$\{ (k, d) \mid 1 \leq k < \infty , d | (k,N) \}$$

Now, $$\{ (k, d) \mid 1 \leq k < \infty , d | (k,N) \}=\{ (k, d)| 1 \leq k <\infty , d | k, d |N \}\\ =\{ (k, d)| d |N, 1 \leq k <\infty , d | k \}$$

Thus $$\sum_{k=1}^{\infty}\left(\mu(d)\right)\frac{t^k}{k}\\ =\sum_{ (k, d)| d |N, 1 \leq k <\infty , d | k }\left(\mu(d)\right)\frac{t^k}{k} \\ =\sum_{ d |N}\sum_{ 1 \leq k <\infty , d | k }\left(\mu(d)\right)\frac{t^k}{k} \\ =\sum_{ d |N}\left(\mu(d)\right) \left(\sum_{ 1 \leq k <\infty , d | k }\frac{t^k}{k}\right) $$