Counterexample to show the set$ \left\{(x_{n})\in l^{p}:{\sum_{n=1}^{\infty}}x_{n}=0\right\}$ is not closed in $l^{p}$ Let $1\le p < \infty$ and  $G=\left\{(x_{n})\in l^{p}:\displaystyle{\sum_{n=1}^{\infty}}x_{n}=0\right\}$.
My attempt was to construct a sequence such that its limit exists but it's not in $G$. Using power series I tried to define the following sequence
$$
  x=(x_{n}) 
$$
where $x_{n}=\left(1,-1,\dfrac{1}{4},-\dfrac{1}{4},\cdots,\dfrac{1}{n^{2}},-\dfrac{1}{n^{2}},0,0,\cdots\right)$. Clearly, every $x_{n}\in G$. However, its limit is
$$
  x=\left(1,-1,\dfrac{1}{4},-\dfrac{1}{4},\cdots,\dfrac{1}{n^{2}},-\dfrac{1}{n^{2}},\cdots\right)
$$
But $x$ is in $G$. Is it possible to fix that counterexample or there exists another counterexample?
 A: The result is true only for $p\in(1,\infty]$. Upper index $n$ will denote the $n$th sequence; lower index $k$ will denote the $k$th term of the sequence. Consider first an auxillary sequence
$$x^n = \left(1,-1,\frac12,\frac{-1}2,\dots,\frac1n,\frac{-1}n,0,0,\dots\right)$$
The termwise and $\ell^p$ limit $$x= \left(1,-1,\frac12,\frac{-1}2,\dots,\frac1n,\frac{-1}n,\dots\right) \in \ell^p,$$
and yes, $\sum_{k=1}^\infty x_k = 0$. But $\sum_{k=1}^\infty |x_k| = \infty$. So there is a bijection $\pi:\mathbb N \to \mathbb N$ defining a rearrangement $y$ of $x$, by the formula $y_k = x_{\pi(k)}$, such that
$$ \sum_{k=1}^\infty y_k = 1 \neq 0.$$
Now define the sequence $y^n$ by $y^n_k=x^n_{\pi(k)}$. Clearly $y^n\in G$ and $y^n \to y\in \ell^p$. The result follows.
For $p=1$, we have boundedness $|\sum_{k=1}^\infty x_k | \le \|x\|_{\ell^1}$. So convergence in $\ell^1$ implies convergence of their sums, so necessarily the limit's sum is $0.$
A: For $p\in (1,\infty].$ Let $C$ be the set of all $x=(x_n)_{n\in \Bbb N}$ in $l^p$ such that $f(x)=\sum_{n\in \Bbb N}x_n$ converges. Then $$ (\bullet)\quad \sup_{0\ne x\in C}|f(x)|/\|x\|=\infty.$$ Let $C_0=\{x\in C: f(x)=0\}.$
Then $C_0$ is dense in $C.$ Proof: If $p\in C$ and $f(p)\ne 0$ then for any $\epsilon \in \Bbb R^+$ there exists $q\in C$ with $\|q\|<\min(\epsilon, |f(p)|)$ and $f(q)=-f(p).$ Then $u=p+q\in C_0$ and $\|p-u\|<\epsilon.$
Therefore $\overline C_0\supseteq C\ne C_0$ so $C_0$ is not closed.
Note: The existence of $q$ follows from $(\bullet)$ since $C$  is a vector space and $f$ is linear on $C.$
Note: $(\bullet)$ is false if $p=1.$
Remark: There is much that has been found about certain real normed linear spaces $B$ with vector subspace $C$ and unbounded linear $f:C\to \Bbb R,$ especially when $C$ is dense in $B$. E.g. when  $B$ is some set of real functions and $f(c)$ is the derivative of $c.$ This is not something I know much of.
