I'm reviewing some old exam questions for an upcoming exam and I'm stumped on the following question.
"Does there exist an analytic function on the unit disc with the property that $f(\frac{1} {n}) = \frac{2+n} {3+n^2}$ for $n=2, 3, 4,... $ Give an example or prove it does not exist."
My proof so far is the following: Assume $f$ is analytic. Then as $f$ must be continuous, we define $f(0) = 0$ and can express $f = \sum a_nz^n$.
Let $N$ be the first index such that $a_n \neq 0$. Then $f = a_Nz^N + a_{N+1}z^{N+1} + \ldots = az^Ng(z)$ and where $g(z)$ is analytic and $g(0) = 1$.
Then $|\frac{2+n} {3+n^2}| = |f(\frac{1} {n})| = |a \frac{1} {n^N} g(\frac{1} {n})|$, so $|g(\frac{1} {n})| = |\frac{n^N (2+n)} {a(3+n^2)}|$.
The above expression then tells us that $N = 1$ as otherwise the right hand side would go to infinity as $n$ does. In addition, we necessarily have that $|a| = 1$.
Then $f(\frac{1} {n}) = \frac{a} {n} g(\frac{1} {n})$, so we can find that $g(\frac{1} {n}) = \frac{n^2 + 2n} {an^2 + 3a}$.
As $n \to \infty$, the right hand side goes to $\frac{1} {a}$. However this is not necessarily a contradiction, so I'm stuck. Any advice on how to continue on or advice on a different method would be much appreciated.
Based on identity theorem, the function $g(z)=\dfrac{2z^2+2z}{3z^2+4}$ agrees with $f(z)$ in infinite points with a limit point so $f(z)=g(z)$ on $D$.