Solution: Suppose that for every $n \in \mathbb{N} $, $f\left( \frac{1}{n} \right) = \frac{1}{n+1} $. Consider the function \[ g(z) \coloneqq f(z) - \frac{z}{z + 1}, \quad \vert z \vert < 1. \] Since $f$ is analytic in the unit disc, and $\frac{z}{1 + z}$ is also analytic on the unit disc, so $g$ is analytic in $\vert z \vert < 1$. For any $n\in \mathbb{N} $, we have \begin{align*} g\left( \frac{1}{n} \right) & = f\left( \frac{1}{n} \right) - \frac{\frac{1}{n}}{1+ \frac{1}{n}} \\ & = \frac{1}{n+1} - \frac{1}{n+1} = 0. \end{align*} Since the sequence $\left( \frac{1}{n} \right) $ converges to $ 0 \in D(0,1) = \{ z: \vert z \vert < 1 \} $, therefore, by the uniqueness theorem (identity theorem), we have \[ g \equiv 0 \implies f(z) = \frac{z}{1 + z}, \quad z \in D(0,1). \] Since $f$ is given to be analytic in the closed unit disc $\overline{D(0,1)} $, so the above function should be continuous at $z = -1 \in \overline{D(0,1)} $. But note that for any sequence $\left( z_n \right) $ converging to $-1$, \begin{align*} \lim_{n \to \infty} f\left( z_n \right) = \frac{z_n}{1 - z_n}, \end{align*} which does not exist. Hence, no such analytic function exists.