I am trying to justify the last statement, but I have failed to do it.
If $\sum \epsilon_k < \infty$, then $\epsilon_k \to 0$. Let $a_j = 1-\epsilon_j$. Then, this implies that for $\varepsilon >0$ there exists $N$ such that for $j > N, |a_j - 1| < \varepsilon$. I don't know how to finish from here.
For the opposite direction, suppose that $\lambda_j > 0$. We have that $$\lambda_j = (1- \sum_{k=j}^\infty \epsilon_k + \cdots).$$ If the sum diverges, $\lambda_j < 0$, which is a contradiction. Does it sound okay?