# Is the set precompact in $\ell_1$

I need to understand is the set $$\{x\in \ell_1 :\sum_{n=1}^{\infty} \frac{\lvert x_n \rvert}{\sin\left(\frac{1}{n}\right)} \leq 1 \}$$ precompact in $$\ell_1$$.

I tryed to use criterion: set in $$\ell_1$$ is precompact if and only if it is bounded and $$\forall \varepsilon > 0 \exists N \in \mathbb{N}:\forall x \sum_{n=N+1}^{\infty} \lvert x_n \rvert < \varepsilon$$.

Boundness is obvious.

I had idea to use $$\lvert x_n \rvert \leq \sin\left(\frac{1}{n}\right)$$ but i think it is useless because $$\sum_{n=1}^{\infty}\sin\left(\frac{1}{n}\right)$$ diverges.

Is it precompact or not?

For a sequence $$a_n>0$$ and $$a_n\to 0$$ consider the operator $$Tx=\sum_{k=1}^\infty a_kx_ke_k$$ where $$\{e_k\}$$ denotes the standard basis in $$\ell^1.$$ The operator $$T$$ is compact as the norm limit of finite dimensional operators $$T_nx=\sum_{k=1}^na_kx_ke_k$$ Therefore $$T(B_1)$$ is a precompact set, where $$B_1$$ denotes the unit ball. However $$T(B_1)=\left \{y\in \ell^1\,:\,\sum_{k=1}^\infty a_k^{-1}|y_k|\leq 1\right \}$$ We can apply the above to $$a_n=\sin(n^{-1}).$$ The same reasoning can be performed for any $$\ell^p$$ space, $$1\le p\le \infty.$$
Since $$\sin (x) \le x$$, we have that $${n} \le \frac{1}{\sin(1/n)}$$ for $$n > 0$$.
Let $$\varepsilon > 0$$ be given. We have by Cauchy's inequality that for any $$N > 1$$ \begin{align*} \sum_{n=N}^{\infty}|x_n| &= \sum_{n=N}^{\infty}n\cdot\frac{1}{n}|x_n|\\ &\le \left(\sum_{n=N}^\infty\frac{1}{n^2} \right)^{1/2}\left( \sum_{n=N}^\infty n^2 |x_n|^2 \right)^{1/2} \\ &\le \left(\sum_{n=N}^\infty\frac{1}{n^2} \right)^{1/2}\left( \sum_{n=N}^\infty n|x_n| \right)\\ &\le \varepsilon \sum_{n=1}^\infty \frac{|x_n|}{\sin(1/n)} \le \varepsilon \end{align*} for $$N$$ chosen large enough and independent of $$(x_n)$$. We have used the inclusion of the $$\ell_p$$ spaces (i.e. $$\Vert x \Vert_2 \le \Vert x \Vert_1$$) in passing to the last line. We have shown that the given set satisfies your criterion.