A dense subspace of L^2 Let $\mathcal{H}$ be the Hilbert space of holomorphic functions defined on the unit disc $D\subset\mathbb{C}$ which is the clousure of the complex polynomial functions on the disc with respect to the inner product given by
$\langle f(x),g(x)\rangle:= \int_0^{2\pi}f(e^{i\theta})\overline{g(e^{i\theta})}\dfrac{d\theta}{2\pi}.$
My question is the following:
Why is the span of $\{\dfrac{1}{z-n}\}_{n=2}^{\infty}$ a dense subset of $\mathcal{H}$? 
I will be grateful for any help.  
 A: Polynomials are dense in $\mathcal H$.    So it suffices to show that every 
power $z^m$ can be approximated uniformly on $D$ by linear combinations of $1/(z-n)$.  Use induction on $m$.
 First of all, consider $m=0$.
$$  \eqalign{-\dfrac{n}{z-n}  &= 1 + \sum_{j=1}^\infty \dfrac{z^j}{n^j} = 1 + Q_n(z) \cr
\left|Q_n(z)\right| & \le 1/(n-1) \ \text{on $D$}\cr
& \to 0 \ \text{as}\ n \to \infty\ \text{uniformly on $D$}}$$
Now for the induction step.  For any positive integer $m$,
$$ -\dfrac{n^{m+1}}{z-n} = \sum_{j=0}^\infty n^{m-j} z^j = P(z) + z^m + z^m Q_n(z)$$
where $P(z)$ is a polynomial of degree $m-1$.  By the induction hypothesis,  $P(z)$ can be approximated uniformly on $D$ by linear combinations of $1/(z-n)$, while $z^m Q_n(z) \to 0$ uniformly on $D$, so we conclude that
$z^m can also be approximated in this way.
A: I'll assume that if $f(z) = \sum_{k=0}^{\infty}a_kz^k, g(z) = \sum_{k=0}^{\infty}b_kz^k$ are in this space, then $\langle  f, g \rangle = \sum a_k\bar {b_k}.$
Suppose $f(z)=\sum_{k=0}^{\infty}a_kz^k \perp M,$ where $M$ is the span you described. Note $1/(n-z) = -(1/n)(1+(z/n) + (z/n)^2 + \cdots).$ So then
$$\langle \sum_k a_ke^{ikt}  , \sum_k e^{ikt}(1/n^k) \rangle = \sum_k a_k(1/n)^k =f(1/n) = 0,$$
$n=2,3,\dots.$ That means the zero set of $f$ in the disc has a limit point (namely $0$). As $f$ is holomorphic, that says $f\equiv 0.$
