TITLE: If $\displaystyle \sum_{n=0}^{\infty}c_{n}x^{n}=0$, show that $c_n= 0$ for any $n$
QUESTION [1 upvotes]: Suppose that $f(x)=\displaystyle\sum_{n=0}^{\infty}c_{n}x^{n}$ for all $x$ with the radius of convergence $R>0$. If $\displaystyle\sum_{n=0}^{\infty}c_{n}x^{n}=0$, show that $c_n=0$ for any $n$.

REPLY [3 votes]: Neither differentiability considerations nor a proof by induction are necessary. Instead, assume that some $c_n$ are not zero, then consider $N=\inf\{n\mid c_n\ne0\}$ and note that $f(x)\sim c_Nx^N$ when $x\to0$ (in the sense that $f(x)/x^N\to c_N$ when $x\to0$).
In particular there exists some positive $\varepsilon$ such that $f(x)\ne0$ for every $x\ne0$ such that $|x|\lt\varepsilon$. This contradicts the hypothesis hence $c_n=0$ for every $n$.
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