Question: $(l^2,\|\cdot\|\_2)$ is a Hilbert space with scalar product $\langle x,y\rangle=\sum^{\infty}\_{k=1}x\_ky\_k$. 

How can I show that every vector $x\in l^2$ can be written in a form $\sum^{\infty}\_{k=1}x\_ke^k$ where $e^k,k\in N$ are unit vectors?

Answer: 
By definition equality 
$$
x=\sum\limits\_{k=1}^\infty x\_k e^k
$$ 
means that 
$$
x=\lim\limits\_{n\to\infty}\sum\limits\_{k=1}^n x\_k e^k
$$
which by definition means that 
$$
\lim\limits\_{n\to\infty}\left\Vert x-\sum\limits\_{k=1}^n x\_k e^k\right\Vert\_2=0\tag{1}
$$
Now use the fact that $\Vert z\Vert\_2^2=\langle z,z\rangle$ to reduce $(1)$ to 
$$
\lim\limits\_{n\to\infty}\sum\limits\_{k=n+1}^\infty |x\_k|^2=0\tag{2}
$$ 
Note that I used here the fact that $\Vert e^k\Vert\_2=1$. I suggest you to recall that $x\in \ell\_2$, to understand why $(2)$ holds.
Let's take $x$ and define $x\_k:=\langle x,e\_k\rangle$. Now we need to prove that the vector

$$y=\sum\_k x\_ke\_k$$
exists and equal to $x$.

First, we show that it exists: the series converges by [pythagorean identity](http://en.wikipedia.org/wiki/Hilbert_space#Pythagorean_identity) ); moreover, it converges to the norm of $x$, because of [Parseval's identity](http://en.wikipedia.org/wiki/Parseval_identity).

$$\|y\|^2=\left\|\sum\_k x\_ke\_k\right\|=\sum\_k x\_k^2 = \|x\|^2.$$

Now clearly the difference $y-x$ is orthogonal to the $E$ - the basis of $H$, hence $y-x\in Cl(E)=H$ (because our space is complete), thus $y-x=0$.