I was thinking about proving a set $S$ is compact in $l^{1}$ space.
$S = \{f \in l^{1} : |f(k)| \leq \frac{1}{k} \forall k \}$
where $\{f_{n}\}_{n=1}^{\infty}$ be functions such that $f_{n} : \Bbb{N} \rightarrow \Bbb{R}^{+}$
I thought of applying that a set $A \subset l^{1}$ is compact if and only if $A$ is closed, bounded and $Sup_{x \in A} \sum_{j \geq n}|x_{j}| \rightarrow 0$ as $n \rightarrow \infty$.
Discussing a bit let me think of taking $f(n) = \frac{1}{n}$, but it would not belong to $l^{1}$ as the harmonic series would not converge. How to proceed with this question.
EDIT -
Also, how the result changes if we would consider $l^{\infty}$ instead of $l^{1}$ space. I think in this case we can take for a fixed integer $N$, $v_{k} = \frac{1}{k}$ for $1 \leq k \leq N$ and $v_{k} =0$ otherwise.
then here $||v|| = 1$ and since $N$ is arbitrary so its bounded always, so the set is compact if we change the space from $l^{1}$ to $l^{\infty}$. Is this correct?.
The set $S$ is not bounded in $\ell^1$: for a fixed integer $N$, let $v$ be the element of $S$ define as $v_k=1/k$ for $1\leqslant k\leqslant N$ and $v_k=0$ otherwise. Then $\left\lVert v\right\rVert_1=\sum_{k=1}^N1/k$ hence $\sup_{v\in S}\left\lVert v\right\rVert_1\geqslant \sum_{k=1}^N1/k$. As $N$ is arbitrary and the series $\sum_{k\geqslant 1}1/k$ is infinite, we derive that $S$ is not bounded.
If we consider instead $\ell^\infty$, that is, $$S = \left\{f \in l^{\infty} : |f(k)| \leq \frac{1}{k} \forall k \right\}, $$ then $S$ is closed (since the functional $f\mapsto f(k)$ is continuous for all $k$) and relatively compact: for a fixed $\varepsilon$, choose $K$ such that $1/K\lt\varepsilon$ and use relative compactness of $\left[-1,1\right]^k$.