Let $((X_n, d_n))_{n = 1}^{\infty}$ be a sequence of metric spaces and let $X := \Pi_{n = 1}^{\infty} X_n$. Equipping the product space $X$ with the metric $$ d : X \times X \to [0,\infty), \quad (x,y) \mapsto \sum_{n = 1}^{\infty} \frac{1}{2^n} \frac{d_n (x_n, y_n)}{1 + d_n (x_n, y_n)} $$ makes the pair $(X, d)$ into a metric space. The following characterisation of convergence in $(X, d)$ is well-known.
A sequence $(x_k)_{k = 1}^{\infty} \subset X$ converges to an element $x \in X$ if, and only if, the sequence $(x_n^{(k)})_{k = 1}^{\infty} \subset X_n$ converges to $x_n \in X_n$ for each $n \in \mathbb{N}$.
It is quite surprisingly, but I am not able to find a direct proof of this fact in the literature. The forward implication is readily verified and the converse implication follows from Lebesgue's dominated convergence theorem. However, I am curious whether there is a more direct or standard proof of the converse direction. I do namely not think that the dominated convergence theorem is really needed for this.
Any comment or reference is highly appreciated.
The topology induced by your metric is the same as the usual product topology on the space $\prod_{i = 1}^\infty X_i$ where the topologies on $X_i$ are the ones induced by $d_i$. This fact can be found (with proof of course) on p. 259 in the book General Topology by Ryszard Engelking found here. Then you may use simply that a sequence converges in the product space if and only if every projected sequence converges. This is suggested as exercise 6 in Munkres Topology, p. 118 found here.