Blaschke product

Remarks

Under the condition above, the product converges uniformly on compact subsets of $\mathbb{D}$. So $B$ is holomorphic with $\abs{B} \le 1$, and its zeros are exactly $0$ and the $a_n$, with multiplicity. Moreover the radial limit of $B$ has modulus $1$ at almost every point of the circle, so $B$ is an inner function. The factor $\abs{a_n}/a_n$ makes the $n$-th factor positive at $z = 0$; some authors also allow a constant factor of modulus $1$.

Example: for $a \in \mathbb{D} \setminus \set{0}$, the single factor $B(z) = \frac{\abs{a}}{a} \frac{a - z}{1 - \bar a z}$ is a finite Blaschke product. For $\abs{z} = 1$ one has $\abs{1 - \bar a z} = \abs{\bar z - \bar a} = \abs{a - z}$, so $\abs{B(z)} = 1$ on the circle.

Non-example: the points $a_n = 1 - 1/n$ for $n \ge 2$ fail the summability condition, since $\sum 1/n = \infty$. For them the partial products tend to $0$ at every point of $\mathbb{D}$, so no Blaschke product has these zeros. In fact no bounded holomorphic function other than $0$ vanishes at all of them.