Compact operators on $\ell^1$

Let $$T\in \ell^1$$, $$Tx = (\lambda_1x_1,\dots,\lambda_nx_n,\dots)$$. Want to show that if $$T$$ is compact, then $$\lambda_n\to0$$.

I know for $$p\in(1,\infty]$$, canonical basis $$e_n \rightharpoonup 0$$ (so $$T(e_n)\rightharpoonup 0$$), and $$T(e_n)\in \overline{T(B(0,1))}$$, which is compact, so we can get $$T(e_n)\to 0$$, and $$\lambda_n\to 0$$.

But what should I do with this problem when $$p=1$$, there is no weak convergence to $$0$$. Can someone help me with this? Thanks

Suppose that $$T\colon\ell^1 \to \ell^1$$ is compact. Then the sequence $$\left(Te_n\right)_{n\geqslant 1}$$ admits a subsequence $$\left(Te_{n_k}\right)_{k\geqslant 1}$$ which converges to some $$v$$ (strongly) in $$\ell^1$$. Look at $$\left\lVert Te_{n_{k+1}}-Te_{n_k}\right\rVert_1$$ to conclude that $$\lambda_{n_k}\to 0$$.
Now apply the previous result to $$\left(Te_{N_j}\right)_{j\geqslant 1}$$ for a fixed sequence $$N_j\uparrow \infty$$ instead of $$\left(Te_n\right)_{n\geqslant 1}$$ to see that each subsequence of $$\left(\lambda_n\right)_{n\geqslant 1}$$ admit a further subsequence with converges to $$0$$.