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.  

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 $f\equiv 0.$
