Question: > 
> Let $l^2$ be the space of square summable sequences with the inner product $\langle x,y\rangle=\sum\_\limits{i=1}^\infty x\_iy\_i$.  
> 
>  (a) show that $l^2$ is H Hilbert space. 
> 
> 
> 

To show that it's a Hilbert space I need to show that the space is complete. For that I need to construct a Cauchy sequence and show it converges with respect to the norm. However, I find it confusing to construct a Cauchy sequence of sequences?

Answer: 
Let $(\mathbf{x\_n})$ be a Cauchy sequence in $l^2$, where $\mathbf{x\_n} = (x\_1^{(n)},x\_2^{(n)},\ldots)$, i.e., given $\epsilon >0$ there exists a natural number $N$ such that for all $m,n\geq N$ 
\begin{equation}
\|\mathbf{x\_n}-\mathbf{x\_m}\| = \left(\sum\_\limits{j=1}^{\infty}|x\_j^{(n)}-x\_j^{(m})|^2\right)^{\frac{1}{2}} <\epsilon
\end{equation}
In particular, it follows that for every $j=1,2,\ldots$ we have 
\begin{align}
|x\_j^{(n)}-x\_j^{(m)}| < \epsilon && (m,n\geq N).
\end{align}
That is for each fixed $j$ the sequence $(x\_j^{(n)},x\_j^{(n)},\ldots)$ is a Cauchy sequence in the scalar field $\mathbb{R}$ or $\mathbb{C}$ and hence it converges. Let $x\_j^{(n)} \to x\_j$ as $n \to \infty$. Using these limits now define $\mathbf{x} = (x\_1,x\_2,\ldots)$. Now, with this basic setting try to show that $(\mathbf{x\_n})$ converges to $\mathbf{x}$.
Pervin = Pervin, W. J.: *Foundations of General Topology*, New York: Academic press, 1964.   

See the proof of Pervin, p.119, Theorem 7.1.2.  

All you need to do is replace $H$ in the proof with $l^2$.  

The proof is divided into two steps:  

Step 1: ($\{x\_n\}$ is Cauchy) $\Rightarrow$ ($x\_n\to x$) [Pervin, p.119, l.$-$13--l.$-$1].  

Step 2: $x\in l^2$ [Pervin, p.120, l.1--l.8].