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.

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
$$\lambda_n = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \Lambda(s) n^{s-1} ds.$$
