Is the free product of residually finite groups always residually finite? Suppose groups $G$ and $H$ are residually finite. Does that imply, that $G \ast H$ is residually finite?
What have I tried to prove this:
Suppose, $a = g_1h_1g_2h_2…g_nh_n \in G \ast H$, $g_1, .. g_n \in G$, $h_1, … , h_n \in H$ and $b = g_1g_2…g_n \neq e$, then the natural homomorphism $\alpha: G \ast H \to \frac{G \ast H}{\langle \langle H \rangle \rangle} \cong G$ maps $a$ to $b$. Now suppose, that $\beta$ is the homomorphism from $G$ to a finite group $K$, such that $\beta(b)$ is non-trivial (such homomorphism exists as $G$ is residually finite). Then $\beta \alpha$ is the homomorphism that maps $a$ to a non-trivial element of a finite group. 
The same arguments can be applied in case, when $h_1h_2 …  h_n \neq e$. However, I do not know, what to do in case, when $g_1g_2…g_n = h_1h_2 …  h_n = e$.
 A: Yes. This reduces to the case of a free product of finite groups, which is free-by-finite, and therefore residually finite.
To see this, take your alternating product $a = g_1 h_1 \cdots g_n h_n\neq 1$ in $G\ast H$, and choose normal subgroups $N$ and $K$, of finite index in $G$ and $H$, respectively, such that $a_1 , a_2 ,\ldots , a_n\not\in N$ and $b_1, b_2, \ldots, b_n \not\in K$. (This can be done since, by definition, there are normal subgroups of $G$ excluding each of the $a_i$, and then their intersection excludes all of them and still has finite index in $G$.  Likewise for the $b_i$ in $H$.)  Then the natural homomorphisms $G\to \overline{G} = G/N$ and $H\to\overline{H}=H/K$ extend to a homomorphism $\phi : G\ast H \to \overline{G}\ast\overline{H}$ for which $\overline{a} = \phi(a)\neq 1$.  (Indeed, $\overline{a} = \overline{g_1}\overline{h_1}\cdots\overline{g_n}\overline{h_n}$ is a reduced alternating product in $\overline{G}\ast\overline{H}$ as all of the $\overline{a_i}$ and $\overline{b_j}$ are non-trivial in their respective finite factors.)  Since $\overline{G}\ast\overline{H}$ is a free product of the finite groups $\overline{G}$ and $\overline{H}$, it is free-by-finite, so it is residually finite.  Hence, there is a homomorphism $\psi : \overline{G}\ast\overline{H} \to Q$, where $Q$ is finite, such that $\psi(\overline{a})\neq 1$.  Then $\psi(\phi(a))\neq 1$, and $\psi\circ\phi$ is a homomorphism from $G\ast H$ to the finite group $Q$ for which $(\psi\circ\phi)(a)\neq 1$.
(There is a tiny corner case to tidy up, where $a\in G\cup H$ but, as $G$ and $H$ are each homomorphic images of $G\ast H$, this is easy to handle.)
