Show that $\lbrace 1, \alpha^k, \alpha^{2k}, \cdots\rbrace$ span $\ell^2$ for $0<|\alpha|<1$ and $k \geq 1$ I'm having troubles to show that if $0<|\alpha|<1$ then the elements $f_k=\lbrace 1, \alpha^k, \alpha^{2k}, \alpha^{3k}, \cdots \rbrace$ span $\ell^2$ for $k \geq 1$. I know I should to use the Vandermonde matrix and its properties, however I don't really know how to proceed.
Can you provide me with some hints?
 A: To clarify, not every element of $\ell^2$ is the sum of finitely many $f_k$; take an element of $\ell^2$ that does not decay as $e^{-tn}$ for some $t$, for example. It is true, however, that the closed span of the $f_k$ (i.e., the closure of the vector space they generate) is $\ell^2$ itself.
Fix $x = (x_0, x_1, \dots)\in \ell^2$, and assume without loss of generality that each $x_i$ lies in $X = [-1, 1]$. Since $x_i \to 0$, there exists (e.g., by the Tietze extension theorem) some continuous $f:X \to X$ with $f(\alpha^n) = x_n$ for each $x$. Fix $\epsilon > 0$, and choose $N$ such that
$$\sum_{n > N} |x_n|^2 < \epsilon.$$
By the Stone-Weierstrass theorem, there exists some polynomial $g(z) = \sum a_n z^n$ with $|g - f| < \epsilon$ on $X$. Then $\xi = \sum a_k f_k\in \ell^2$ has 
$$|\xi_n - x_n| = \left|\sum_k a_k \alpha^{nk} - x_n\right| = |g(\alpha^n) - x_n| < \epsilon$$
for all $n$. Now bound the $\ell^2$-norm of $\xi - x$, using the fact that the $(f_k)_n$ decay exponentially in $n$.
A: Recall that a set $S\subseteq \ell^2$  is dense if $({\bf c},s)=0$ for all $s\in S$ implies ${\bf c}=0$.
Fix $\alpha$ such that $0<|\alpha|<1$ and let 
 $$S=((1,\alpha^k,\alpha^{2k},\dots):k=1,2\dots)\subset \ell^2.$$ 
Let ${\bf c}\in \ell^2$ be such that $({\bf c},s)=0$ for all $s\in S$ and define 
$$f(x) = ({\bf c},(1,x,x^2,\dots)) = \sum_{j=0}^\infty c_j x^j.$$ 
Then $f$ is analytic on $(-1,1)$ (and by extension on the open unit ball in the complex plane). Since by assumption $f(\alpha^k)=0$ for all $k=1,2,...$, and $\alpha^k \to 0$ as $k\to\infty$, it follows from the uniqueness theorem for analytic functions that $f$ is identically zero. Therefore, $c_k = \frac{f^{(k)}(0)}{k!}=0$, proving that ${\bf c}=0$. Thus $S$ is dense.   
