sequences and series - What's the closed form of this :$sum_{n=1}^{+infty}frac{(-1)^nphi(n)}{n}$

I have checked some links related the below sum which is related to The Euler totient function to check if it has any known closed form but i don't find anything then my question here is :

Question:
What is the closed form of this :$\sum_{n=1}^{+\infty}\frac{(-1)^n\phi(n)}{n}$ , where $\phi(n)$ is Euler totient function ?

Answer

As stated by reuns in the comments, for any $s$ with a large enough real part we have

$$ \sum_{n\geq 1}\frac{\varphi(n)}{n^s} = \prod_{p}\left(1+\frac{\varphi(p)}{p^s}+\frac{\varphi(p^2)}{p^{2s}}+\frac{\varphi(p^3)}{p^{3s}}+\ldots\right)= \prod_{p}\frac{p^s-1}{p^s-p}$$
by Euler's product, hence
$$ \sum_{n\geq 1}\frac{\varphi(n)}{n^s} = \prod_p \frac{1-\frac{1}{p^{s}}}{1-\frac{1}{p^{s-1}}}=\frac{\zeta(s-1)}{\zeta(s)}$$
$$ \sum_{\substack{n\geq 1\\n\text{ odd}}}\frac{\varphi(n)}{n^s} = \prod_{p>2} \frac{1-\frac{1}{p^{s}}}{1-\frac{1}{p^{s-1}}}=\frac{\zeta(s-1)}{\zeta(s)}\cdot\frac{2^s-2}{2^s-1}$$
$$ \sum_{n\geq 1}\frac{(-1)^n \varphi(n)}{n^s}=\frac{\zeta(s-1)}{\zeta(s)}\left(1-2\cdot\frac{2^s-2}{2^s-1}\right) =-\frac{\zeta(s-1)}{\zeta(s)}\cdot\frac{2^s-3}{2^s-1}$$
but the series in the LHS is convergent only for $\text{Re}(s)>2$.