Show that $l^2$ is a Hilbert space 
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?

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 $\mathbf{x}$.
