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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.

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really sorry for the edit. accidentally had $\alpha^n$ in the summation instead of $\alpha_n$ I'm glad you realised that Richard. Sorry xaviermo2 –  FBI Nov 17 '12 at 12:40

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up vote 4 down vote accepted

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 $n\ge 1$.

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How did you get that $f(1) = 0$? –  Matt N. Nov 17 '12 at 12:34
    
Thank you very much. I was trying to consider a power series but (stupidly) couldn't see which sequence $z_n$ to take so I could apply the identity theorem. –  FBI Nov 17 '12 at 12:35
    
$ f(1) = 0 $ as $ f(1) = f(1/1) $ and each $f(1/k) = 0$ –  FBI Nov 17 '12 at 12:36
    
@FBI: You are welcome. By the way, there is a typo in your question which puzzles others: $\alpha^n$ should be $\alpha_n$. –  23rd Nov 17 '12 at 12:38
    
@MattN.: There was a typo in FBI's question. Now it is corrected. –  23rd Nov 17 '12 at 12:40

Following Xavier's excellent hint:

$$\alpha\neq 0\;\;,\;\;k\in\Bbb N\Longrightarrow \sum_{n=0}^\infty\frac{\alpha^n}{k^n}=\sum_{n=0}^\infty\left(\frac{\alpha}{k}\right)^n=\begin{cases}\text{diverges}&\text{if}\,\;\;\left|\frac{\alpha}{k}\right|\geq 1\\{}\\\frac{1}{1-\frac{\alpha}{k}}=\frac{k}{k-\alpha}&\text{if}\,\;\;\left|\frac{\alpha}{k}\right|<1\end{cases}\;\;\neq 0$$

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The answer above refers to the original post, with $\,\alpha^n\,$ was in the sum and there was nothing about $\,\alpha_n\,$ in it –  DonAntonio Nov 17 '12 at 13:02

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