Final answer:
The given sequence {xn} with terms (0, ..., 0, √n, √n, 0, 0, ...) does not converge weakly in ℓ¹ because the terms grow without bound (√n), which prevents stabilization required for weak convergence, and the sequence is not bounded in the ℓ¹ norm.
Step-by-step explanation:
The question posed involves examining the convergence of a sequence in ℓ¹ (the space of absolutely summable sequences). Specifically, it asks whether the sequence {xn}[infinity] n=1 with xn represented as (0, ..., 0, √n, √n, 0, 0, ...) converges weakly in ℓ¹. In ℓ¹, a sequence {xn} is said to converge weakly to some element x if for every bounded linear functional f on ℓ¹, the limit of f(xn) as n goes to infinity equals f(x).
However, the given sequence does not converge weakly in ℓ¹. The reason is that for any typical bounded linear functional on ℓ¹, which acts by summing the product of the sequence elements with a fixed sequence from the dual space, the sequence will contain terms that grow without bound (√n). This growing term implies that the linear functional applied to these sequences will also not stabilize at a finite limit, which is a requirement for weak convergence.
Furthermore, for the weak convergence in ℓ¹, the sequence would need to be bounded in the ℓ¹ norm, which it is not due to the √n term. Therefore, the given sequence does not converge weakly in ℓ¹.