There exists no function $f$ that is analytic in the unit disk $\Delta $ such that $$f(1/n)=f(-1/n)=n^{-2k+1}\quad \forall n\geq 2$$ where $k \in \mathbb{N} $ is fixed.
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Remember that the zeroes of a nontrivial analytic function must be isolated. – Jared Apr 17 '13 at 01:54
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Apply the isolated zero principle twice: if the zero set of an analytic function on the connected open set $\Delta$ has a limit point in $\Delta$, then $f$ is zero on $\Delta$.
See here for a proof by Andres Caicedo.
Assume such a function exists and consider the limit point $0=\lim_{n\rightarrow +\infty}\frac{1}{n}$ in the following two cases.
From the first equality, the function $f(z)-f(-z)=0$ on $\Delta$, i.e. $f$ is even.
From the second equality, we get that $f(z)-z^{2k-1}=0$ on $\Delta$, i.e. $f(z)=z^{2k-1}$ is odd.
So $f$ is both odd and even, i.e. $f(z)=0$ on $\Delta$. Contradiction with the second equality.
