There was a question in my exam :
Let $x=(x_n)=(0,0,..,0,1/n^2,0,0,..)\in c_{00}$
Show that $\sum_{n=1}^n x_n$ is absolute convergent but not convergent in $c_{00}$ space wrt $\|.\|_{\infty}$ norm.
The reason of my confusion can I use terms that include "$n$" in components of vectors?
I have said it is absolute convergent since
$\sup_{n\in \mathbb N}|x_n|=1/n^2$ and $\sum_{n=1}^n 1/n^2 \lt \infty$
for not being convergent I have said,
Let $S_j=(0,0,..,1/n^2)$ is partial sum of $x_n$
$S=(1/n,1/n,...,1/n)$
$\sup_{j\in \mathbb N}|S_j-S|=1/n \lt \varepsilon$
$S_j \to S$ and $S \notin c_{00} $
Thus $S_j$ sequence is not convergent in $c_{00}$
I have written more detailed in exam, I know writing is not well enough in here but I am pledge in StackExchange :)
As I said before I confused about using "$n$" in components. For writing $S$, I'm inspired from $(0,0,..,0,1/n^2,0,0,..)$.(since it has a component which has "$n$"). Is $S=(1/n,1/n,...,1/n)$ incorrect notation?
I have trouble understanding your notation, or you seem to have calculated the partial sums wrong.
If we denote the sequence of partial sums by $(S_n)_n$ then
$$S_n = \sum_{i=1}^n x_i = \left(1, \frac1{2^2}, \frac1{3^2}, \ldots, \frac{1}{n^2}, 0, 0, \ldots\right)$$
Now recall that convergence in $\|\cdot\|_\infty$ implies coordinate-wise convergence, in the sense that if $x_n \xrightarrow{\|\cdot\|_\infty}x$ then the sequence of $i$-th coordinates of $(x_n)_n$ converges to the $i$-th coordinate of $x$, for every $i \in \mathbb{N}$.
Hence the only possible candidate for $\lim_{n\to\infty} S_n$ is its coordinate-wise limit, which is
$$\left(\frac1{n^2}\right)_n = \left(1, \frac1{2^2}, \frac1{3^2}, \ldots, \frac{1}{n^2}, \frac{1}{(n+1)^2}, \ldots\right)$$
However, this is not an element of $c_{00}$ so $(S_n)_n$ cannot converge.