convergence of sequence. Let $l^p=\displaystyle \bigg\{\langle x_k\rangle \in \mathbb R^n | \sum^{\infty}_{k=1} |x_k|^p <\infty \bigg\}$
Let $e^k, k\in \mathbb N$ be given by:  
\begin{align*}
e^1&=\langle 1, 0, 0,\ldots,0\ldots\rangle \\
e^2&=\langle 0, 1, 0,\ldots,0\ldots\rangle \\
\vdots & \\
e^n&=\langle 0,\ldots,0,1,0,\ldots \rangle  
\end{align*}
Show that for all $x\in l^p$
$\displaystyle x=\sum^{\infty }_{k=1} x_k e^k$  
If $\displaystyle x=\sum^{\infty }_{k=1} x_k e^k$, I have to show that $\displaystyle x=\lim_{n \to \infty }\sum^{n}_{k=1} x_k e^k$.
Let write $S_n$ as $S_n=\sum^{n}_{k=1} x_k e^k=\langle x_1, x_2,\ldots,x_n,0,\ldots\rangle$  
I have to show that $(\forall\epsilon >0)($ there is $n_0\in N)(\forall n\geq n_0)(||x-S_n||_p <\epsilon)$, where $\displaystyle \|x-S_n\|_p=\sqrt[p]{\sum^{\infty}_{k=1}|x_k-S^{k}_{n}|^p}$.
 A: Define $\displaystyle \ell^p = \bigg\{ \langle x_n \rangle \in \mathbb R : \sum_{k = 1}^\infty \vert x_k \vert^p < \infty \bigg\}$ and $e_k := \{0, \ldots, 0, \underbrace{1}_{k^{\text{th}} \text{coordinate}}, 0, \ldots\}$.
We want to show that $x = \displaystyle \sum_{k = 1}^\infty x_k \cdot e_k = \lim_{n \to \infty} \sum_{k = 1}^n x_k \cdot e_k$ where $x = \langle x_k \rangle \in \ell^p$.
Let $\varepsilon > 0$. Since $x \in \ell^p$, then $\displaystyle \sum_{k = 1}^\infty \vert x_k \vert^p < \infty$ which means that the tail end goes to zero, that is, there exists $N$ such that if $n > N$, then $\displaystyle \sum_{k = n}^\infty \vert x_k \vert^p < \varepsilon^p$.
If $n > N$, then
\begin{align*}
\bigg \|\sum_{k = 1}^{n - 1} x_k \cdot e_k - x \bigg \| &= \| \langle x_1, x_2, \ldots, x_{n - 1}, 0, 0, \ldots \rangle - \langle x_1, x_2, \ldots \rangle \| \\
&= \| \langle x_n, x_{n + 1}, \ldots \rangle \| \\
&= \bigg(\sum_{k = n}^\infty \vert x_k \vert^p\bigg)^{1/p} \\
&< \varepsilon.
\end{align*}
Conclude that $\displaystyle \sum_{k = 1}^\infty x_k \cdot e_k = \lim_{n \to \infty} \sum_{k = 1}^\infty x_k \cdot e_k = x$ as desired.
