Let $f(n)$ be the number of prime divisors of a number $n$ counted with their multiplicities. Show that the series $\sum_{n=1}^{+\infty}\frac{(-1)^{f(n)}}{n}$ converges and has sum $0$.
Attempt : The infinite product $$\prod_{p\text{ prime}}\left(1+\frac1p\right)^{-1}$$ diverges to zero.
and we have, $$\left(1+\frac1p\right)^{-1}=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{p^{k}}$$
Formally we can write : $$\prod_{p\text{ prime}}\sum_{k=0}^{\infty}\frac{(-1)^{k}}{p^{k}}=\sum_{n=1}^{\infty}\frac{(-1)^{f(n)}}n$$
Unfortunatly I do not see how can I make this rigourous.
I'm afraid there is no simple rigorous proof. The situation is the same as proving $$\sum_{n=1}^\infty\frac{\mu(n)}{n}=0,$$ where $\mu(n)$ is the Möbius function ($\mu(1)=1$; $\mu(n)=(-1)^s$ if $n$ is a product of distinct primes $p_1,\ldots, p_s$ and $\mu(n)=0$ otherwise). For this one needs to use inverse formula for the corresponding Dirichlet series $$\sum_{n=1}^\infty\frac{\mu(n)}{n^s}=\frac{1}{\zeta(s)}$$ (such as Perron's formula). This is as hard as the prime number theorem! (In fact, even equivalent to it.) Detailed proof you can find, for example, in the book Titchmarsh E.C. The theory of the Riemann zeta-function (section 3.13).
The same can be done for the Dirichlet series generating function of your sequence: $$\sum_{n=1}^\infty\frac{(-1)^{f(n)}}{n^s}=\frac{\zeta(2s)}{\zeta(s)}$$ and you'll get the desired proof.
Another approach is to use the following result.
Theorem. Suppose that the series $$F(s)=\sum_{n=1}^\infty\frac{a_n}{n^s}$$ is absolutely convergent for $\Re s>0$, $F(s)$ is regular for $s=0$ and $a_n=O(\frac1n)$. Then the series $$\sum_{n=1}^\infty a_n$$ is convergent and has sum $F(0)$.
(See the book: R. Ayoub, An Introduction to the Analytic Theory of Numbers, chapter II, problems 29-31.) Proving this theorem isn't much easier though.