I have problems to understand the notation of sequences of sequences and limits
Let $(x^j)\subset{l^1}$ be a sequences of sequences given by $x^j_n:=\begin{cases} 0 & n=j \\ \frac{1}{n^2} & \text{else} \end{cases}$
and let be the limit $x_n=\frac{1}{n^2}$ then the limit in $l^1$-norm given by $\sum^\infty_{n=1}|x^j_n-x_n|=\frac{1}{j^2}\rightarrow0$ as $j\rightarrow\infty$.
Is it correct that we compare like this $|x^j_n-x_n|=(1,\frac{1}{4},\frac{1}{9},\cdots,\frac{1}{n^2},0,\frac{1}{(n+2)^2},\cdots)-(1,\frac{1}{4},\frac{1}{9},\cdots,\frac{1}{n^2},\frac{1}{(n+1)^2},\frac{1}{(n+2)^2},\cdots)$
if so I can not see how the tail of the sum is given by $\frac{1}{j^2}$ and not $\frac{1}{(1+n)^2}$ where $j=n+1$ or can we simply substitute $n=j-1$?
Pleas tell if something is unclear
You have a bit of confusion about the notation it seems. You have $$ x^j = \left(1, \frac 14, \ldots, \frac1{(j-1)^2}, 0, \frac1{(j+1)^2}, \ldots\right). $$ Note that the "missing" index is $j$. Note that when we write $x_n^j$ and $x_n$, they are not the whole sequences, but only the $n$'th index. What you have is $$ x - x^j = \left(1, \frac 14, \ldots \right) - \left(1, \frac 14, \ldots, \frac1{(j-1)^2}, 0, \frac1{(j+1)^2}, \ldots\right) = \left(0, \ldots, 0, \frac1{j^2}, 0, \ldots\right), $$ so $$ |x - x^j| = \frac1{j^2}. $$ In index notation, $$ x_n - x_n^j = \begin{cases} \frac1{j^2} & n= j \\ 0 & n\ne j\end{cases}, $$ and then you sum $|x_n - x_n^j|$ over all $n$ to get $\frac1{j^2}$. Remember also, $j$ is fixed here and $n$ is the one we sum over. You can't declare $j=n+1$ or anything like that.
Knowing this, we can now let $j\to\infty$, and we see that $|x - x^j| = \frac1{j^2}\to 0$, proving the convergence. Makes sense?