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?