Existence of an analytic function by estimate

Question

Does there exist an analytic function on the unit disc s.t. $|f\left(\frac1n\right)−\frac{(−1)^n}{n^2}|<\frac{1}{n^3}$ for all $n\geq2$?

See my idea but I don’t know how to complete the proof. I know such analytic function doesn’t exist.

Answer 1

I will like to know if my idea is correct and how to get a contradiction from the hypothesis.

Answer 2

$f$ must be non-constant (why?), so we may write $f(z) = z^m \cdot g(z)$
for some non-negative integer $m$ and analytic $g$ which is non-zero in $B\big(0,\delta\big)$, i.e. a sufficiently small $\delta\gt 0$ neighborhood of zero.

$S:= \big\{\frac{1}{2},\frac{1}{3},\frac{1}{4},\cdots \big\}$

For $z\in S$: application of reverse triangle inequality gives
$\Big \vert z^m \cdot \big\vert g(z)\big \vert − z^2 \Big \vert\lt z^3$

1.) if $m\in\big\{0,1\big\}$ dividing each side by $z^m$ implies $g(0)=0$ which is impossible

2.) if $m\gt 2$, then re-scaling each side by $z^{-2}$ yields
$1 - z^{m-2} \cdot \big\vert g(z)\big \vert \leq \Big \vert z^{m-2} \cdot \big\vert g(z)\big \vert − 1 \Big \vert\lt z$
$\implies 1\leq 0$ by taking limits, which is impossible.

3.) it remains to consider $m=2$.
Multiplying each side by $z^{-2}$ (i.e. multiplying by $n^2$), the original inequality reads
$\big \vert g\left(\frac1n\right)− (−1)^n\big \vert = \big \vert n^2\cdot f\left(\frac1n\right)− (−1)^n\big \vert<\frac{1}{n}$
which is impossible since it implies $-1=1$, by taking a separate limit over odd natural numbers vs even natural numbers