I'm stuck doing the following exercise:
In space $l_2$ consider the following sequence: $$ x_k=\left(1,\frac{1}{2^k},\frac{1}{2^{2k}},\frac{1}{2^{3k}},\dots \right). $$ Show that the space generated by this sequence is dense in $l_2$.
My idea is to show that the only vector orthogonal to each $x_k$ is zero vector (clearly this implies the density). So let $y=(y_0,y_1,\dots)\in l_2$ be orthogonal to each $x_k$ then $$ \sum_{i=0}^{\infty}y_i \dfrac{1}{2^{ik}} =0. $$ Therefore I have a sort of infinite system of linear equation (with $y_i$ are unknown) which is given by intinite Vandermonde matrix. How could I show that the kernel of that matrix is $0$ ? (I guess this will imply that $y=0$ is the only solution).
Define the function $$f\colon x\mapsto\sum_{n=0}^{+\infty}y_nx^n.$$ Since the sequence $\left(y_n\right)_{n\geqslant 0}$ is bounded, the function $f$ is well-defined on $(-1,1)$. Moreover, the function $f$ is continuous on $[-1/2,1/2]$. The assumptions give that $f\left(2^{-j}\right)=0$ for each positive integer $j$. This implies that $f(0)=0$ hence $y_0=0$. Now define $$f_1(x)\colon x\mapsto \sum_{n=0}^{+\infty}y_{n+1} x^n.$$Observe that $f_1\left(2^{-j}\right)=0$ for each positive integer $j$ and that $f_1$ is continuous to get that $y_1=0$. Now, you can show by induction on $n$ that $y_n=0$ for each $n$.
If you know power series, a shorter argument can be given.