Closed set in $\ell^1$ 
Show that the set $$ B = \left\lbrace(x_n) \in \ell^1 : \sum_{n\geq 1} n|x_n|\leq 1\right\rbrace$$
  is compact in $\ell^1$.
  Hint: You can use without proof the diagonalization process to conclude that every bounded sequence $(x_n)\in \ell^\infty$ has a subsequence $(x_{n_k})$ that converges in each component, that is $\lim_{k\rightarrow\infty} (x_{n_k}^{(i)})$ exists for all i.
  Moreover, sequences in $\ell^1$ are obviously closed by the $\ell^1$-norm.

My try: Every bounded sequence $(x_n) \in \ell^\infty$ has a subsequence $(x_{n_k})$
that converges in each component. That is $\lim_{k\rightarrow\infty} (x_{n_k}^{(i)})$ exists for all i.  .And all sequences in $\ell^1$ are bounded in $\ell^1$-norm.
I want to show that every sequence $(x_n) \in B$, has an Cauchy subsequence.
Choose an N and M such that for $l,k > M$ such that $|x_{n_k}^{(i)} - x_{n_l}^{(i)}| < \frac{1}{N^2}$ Then 
$$\sum_i^N |x_{n_k}^{(i)} - x_{n_l}^{(i)}| + \sum_{i = N+1} ^\infty |x_{n_k}^{(i)} - x_{n_l}^{(i)}| \leqslant \frac{1}{N} + \frac{1}{N+1}  \sum_{i = N+1} ^\infty i|x_{n_k}^{(i)} - x_{n_l}^{(i)}| \leqslant \frac{3}{N+1}$$ 
It feels wrong to compine $M,N$ like this, is it? what can I do instead?

We can use and show the following:

Let $K\subset \ell^1$. This set has a compact closure for the $\ell^1$ norm if and only if the following conditions are satisfied:

*

*$\sup_{x\in K}\lVert x\rVert_{\ell^1}$ is finite, and

*for all $\varepsilon>0$, we can find $N$ such that for all $x\in K$, $\sum_{k\geqslant N}|x_k|<\varepsilon$.


These conditions are equivalent to precompactness, that is, that for all $r>0$, we can find finitely many elements $x^1,\dots,x^N$ such that the balls centered at $x^j$ and for radius $r$ cover $K$.
