Show that sequences in $\ell^1$ containing finite number of non-zero elements are dense in $\ell^1$. Take sequences from $\ell^1$ which have a finite number of elements other than $0$. Prove, that these sequences are dense in $\ell^1$. What does it imply?
I  tried to use the definition of densitiy such that 
A is dense in $\ell^1$, if
$$ \forall x\in \ell^1 ,\quad \forall\epsilon>0,\quad \exists a\in A:\quad  d(x,a)<\epsilon$$
Unfortunately I fail to see how this proves the initial assumption.
Any help or suggestion is appreciated!
 A: For $a=(a_n )_{n\in\mathbb{N}}\in \ell_1$ and $q_n (a) =(a_k)_{k=1}^n $ we have $$d(q_n (a) ,a) =\sum_{j=n+1}^{\infty} |a_j | \to 0$$ as $n\to\infty .$
A: Expanding on my comment, but deviating in notation:
$\ell^1$, the space of all summable sequences has the norm $$\|x\|_1 = \sum_{k=1}^\infty x_n,$$
where $x=(x_1,x_2,\dots)$ is a sequence. Hence, the induced metric is given by $$d(x,y)=\|x-y\|_1= \sum_{k=1}^\infty |x_n - y_n|,$$
i.e. the series of the sum of absolute values of the componentwise difference of $x$ and $y$ (you can think of $x$ as a row vector with an infinite number of rows and the additional property that the absolute values of the row entries are summable).
If you want to have $y$ close to $x$, then their difference needs to be a series whose entries sum to a small number.
To get back to your question: How do we approximate any given $x \in \ell^1$ with elements from $A$, the space of sequences which are eventually zero?
For a given $x= (x_1,x_2, x_3, \dots)$, we can 'truncate' $x$ at the point $k$, i.e. we let 
$$x^{(k)}=(x_1,x_2, x_3,\dots,x_k, 0,0,\dots)$$
so that up to the $k-$element, the sequence $x^{(k)}$ coincides with $x$ and beyond the $k-$th element, it's equal to zero.


*

*What's the value of $d(x,x^{(k)})$?

*Why does this value decrease as $k$ increases?

