proofs of $n\sum_{d|n}\frac{(-1)^{d+1}}{d}=\sum_{d|n,d \text{ odd}}d$ I'm looking for other proofs of the identity :

For all integer $n\in \mathbb{N}^{+}$:
  $$n\sum_{d|n}\frac{(-1)^{d+1}}{d}=\sum_{d|n,d \text{ odd}}d $$
  where in the first sums is taken over all dividors of $n$, and the second over all odd divisors of $d$

I have this poof, let $n=2^k(2q+1)$ :
$$\begin{align}n\sum_{d|n}\frac{(-1)^{d+1}}{d}&=&&n\sum_{d|2q+1}\sum_{ i=0} ^k \frac{(-1)^{2^id+1}}{2^id} \end{align}$$
because every divisor of $n$ can be written uniquely as $2^id$ where $d|2q+1$ and $0\leq i \leq k$
$$\begin{align}n\sum_{d|n}\frac{(-1)^{d+1}}{d}&=&&n\sum_{d|2q+1}\left(\frac{1}{d} -\sum_{ i=1} ^k \frac{1}{2^id}\right)\\
&=&&\sum_{d|2q+1}\frac{n}{2^kd} \\
&=&&\sum_{d|2q+1} \frac{2q+1}{d}\\
&=&&\sum_{d|n,d \text{ odd}}d \end{align}$$
So i want to see if there is others proofs, mainly using generating functions because I think that they are powerful enough to solve this sorts of identities. Any hints, suggestions will be greatly appreciated.
 A: Following the suggestion from the comment we can prove algebraic equality as follows.
We have that
$$\sum_{n\ge 1} \frac{(-1)^{n+1}/n}{n^s}
= \sum_{n\ge 1} \frac{(-1)^{n+1}}{n^{s+1}}
= \left(1-\frac{2}{2^{s+1}}\right)\zeta(s+1)
= \left(1-\frac{1}{2^{s}}\right)\zeta(s+1).$$
Therefore
$$\sum_{n\ge 1} \left(\sum_{d|n} \frac{(-1)^{d+1}}{d}\right) \frac{1}{n^s} =
\zeta(s) \left(1-\frac{1}{2^{s}}\right)\zeta(s+1).$$
and the Dirichlet generating function of the LHS is
$$\zeta(s-1) \left(1-\frac{1}{2^{s-1}}\right)\zeta(s).$$
On the other hand we also have
$$\sum_{n\ge 1, \; n\;\mathrm{odd}} \frac{n}{n^s}
= \prod_{p,\; p\;\mathrm{odd}} \frac{1}{1-p^{-(s-1)}}
= \left(1-\frac{1}{2^{s-1}}\right)\zeta(s-1).$$ 
Therefore the Dirichlet generating function of the RHS is
$$\zeta(s) \left(1-\frac{1}{2^{s-1}}\right) \zeta(s-1)$$ 
and we have equality.
Remark. If we want to be rigorous about it, we have equality of the coefficients of the Dirichlet series in the intersection of the two half-planes of convergence because for the Dirichlet series $\Lambda(s) = \sum_{n\ge 1} \frac{\lambda_n}{n^s}$ we have
$$\lambda_n = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \Lambda(s) n^{s-1} ds.$$
