Let $B \subseteq l^2$, $B=\left\{x\in l^2:\sum_{n\geq1}n|x_n|^2\leq1\right\}$, show that $B$ is compact.
My thought: $B$ is closed in $l^2$ which is complete. Then $B$ is complete. It suffices to show $B$ is totally bounded. I think we need to first get rid of the infinite tail sum, i.e., bound all sequences with balls centered at sequences that only have finitely many terms. And then find a ball cover for the finite sequences. But I don't know how to bound the tail sum and I'm stuck.
Theorem. Let $p\in[1,+\infty]$ and $M\subseteq\ell_p$ be a bounded subset such that $$ \lim\limits_{N\to\infty}\sup\{\Vert (0,0,\ldots,0,x_N,x_{N+1},\ldots) \Vert_p:x\in M\}=0 \tag{1}$$ then $M$ is totally bounded.
For the proof, see an answer of that question: How to show that this set is compact in $\ell^2$.
The theorem provides a characterization of totally bounded subsets of $\ell_p$.
Let’s apply it to $B$.
For $x\in B$, we have
$$\Vert x\Vert_2 = \sum_{n\ge1} \vert x_n \vert^2 \le \sum_{n\ge1} n\vert x_n \vert^2 \le 1$$ proving that $B$ in included in the closed ball centered on the origin with radius equal to one. Hence $B$ is bounded.
We’ll be done if we prove that $B$ satisfies condition $(1)$ of theorem above.
For $x \in B$
$$N\sum_{k\ge N} \vert x_k\vert^2 \le \sum_{k\ge N} k\vert x_k\vert^2 \le \sum_{k\ge 1} k\vert x_k\vert^2 \le 1$$ Therefore $$\sup\{\Vert (0,0,\ldots,0,x_N,x_{N+1},\ldots) \Vert_2:x\in B\} \le 1/N$$ and condition $(1)$ is satisfied.
$B$ is totally bounded. And also complete as being a closed subset of a complete space.
Finally $B$ is compact.
