Identity Theorem Let $(\alpha_n)$ be a sequence of complex numbers such that $$ \sum_{n=1}^\infty \frac{\alpha_n}{k^n} = 0 $$ for all $k = 1, 2, 3, ...$ Prove that $\alpha_n=0$ for all $n$
As far as a solution goes, really not sure where to start. Any hints would be appreciated. Thanks.

Consider the power series $f(z)=\sum_{n=1}^\infty\alpha_nz^n$. From $f(1)=0$ you will know that $\lim_{n\to\infty}\alpha_n=0$, so the radius of convergence of the power series is at least $1$. That is to say, $f$ is a well defined holomorphic function on $\mathbb{D}:=\{z\in\mathbb{C}:|z|<1\}$. Your assumption says that $f(\frac{1}{k})=0$ for every $k\ge 1$, but $\lim_{k\to\infty}\frac{1}{k}=0\in \mathbb{D}$, which implies that $f$ must be identically $0$ on $\mathbb{D}$, i.e. $\alpha_n=0$ for every $n\ge 1$.
