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 $\mathbf{x}$.
Hint : Let $(x^n)$ be a Cauchy sequence. We can write $x^n=\sum \_k x\_k^ne\_k$ where $(e\_k)$ is the canonical basis. Show $|x\_k^n-x\_k^m|\leq \|x^n-x^m\|$ so $(x\_k^n)\_n$ is a Cauchy sequence in $\mathbb{C}$, so there exist $x\_k$ such that $\lim x\_k^n=x\_k$ for every $k$. Define $x=\sum x\_ke\_k$ and prove that $\|x^n-x\|\to 0$.