u 2 = 20/ 9 .

a) \({u_2} = {u_1} + \frac{2}{3} = 2 + \frac{2}{3} = \frac{8}{3}\).

b) vn = un + 1 – un Þ \({v_n} = \frac{2}{{{3^n}}}\) \( \Rightarrow {v_2} = \frac{2}{{{3^2}}} = \frac{2}{9}\).

c) \(\lim {v_n} = \lim \frac{2}{{{3^n}}} = 0\).

d) Ta có \({v_1} + {v_2} + ... + {v_{n - 1}} = \frac{2}{3}\frac{{1 - {{\left( {\frac{1}{3}} \right)}^{n - 1}}}}{{1 - \frac{1}{3}}} = 1 - \frac{3}{{{3^n}}},\forall n \ge 2\).

mặt khác v1 + v2 + …+ vn – 1 = un – 2, ∀n ≥ 2.

Nên \({u_n} = 3 - \frac{3}{{{3^n}}}\). Do đó \(\mathop {\lim }\limits_{n \to  + \infty } {u_n} = \mathop {\lim }\limits_{n \to  + \infty } \left( {3 - \frac{3}{{{3^n}}}} \right) = 3\).

Đáp án: a) Sai;    b) Đúng;c) Sai;d) Đúng.