Find all analytic functions in the disk $|z-2|<2$ such that $f \left(2+\frac{1}{n} \right) = \frac{1-n}{5n+3} + \sin{\left(\frac{n \pi}{2} \right)}$ Find all analytic functions (or prove that no such exist) inside the disk $|z-2|<2$ that satisfy the following condition:
$$f \left(2+\frac{1}{n} \right) = \frac{1-n}{5n+3} + \sin{\left(\frac{n \pi}{2} \right)}, \ n \in \mathbb{N}$$
For $n=2k$ the expression simplifies quite a bit since $\sin{\left(\frac{n \pi}{2} \right)}= \sin{\left(k \pi \right)} = 0$ so we're left with $$f \left(\frac{4k+1}{2k}\right) = \frac{1-2k}{10k+3}$$
Similarly for $n=4k-3$ and $n=4n-1$ we can simplify the sinus expression to $1$ and $-1$. So we're getting $3$ different expressions for $f\left(2+\frac{1}{n}\right)$ based on whether $n=2k$ , $n=4k-3$ or $n=4k-1$. My idea is to use the identity theorem for any of these two expressions and prove that no such function exists but I can't quite tie it all together.

Your idea is correct, there is also a simpler one: assume that such a function exists. Then using continuity of $f$ we get
$$f(2) = \lim_{n \to \infty} f \left( 2+\frac{1}{n} \right) = \lim_{n \to \infty} \left( \frac{1-n}{5n+3} + \sin \frac{n \pi}{2} \right).$$
But the limit on the right clearly does not exist, which is a contradiction.
