Cho dãy số (un) với u1 = 2; un + 1 = \({u_n} + \frac{2}{{{3^n}}}\), n ³ 1. Đặt vn = un + 1 – un.

a) \({u_2} = \frac{{20}}{9}\).

b) \({v_2} = \frac{2}{9}\).

c) \(\mathop {\lim }\limits_{n \to  + \infty } {v_n} = 2\).

d) \(\mathop {\lim }\limits_{n \to  + \infty } {u_n} = 3\).

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

b) Có v_n = u_{n + 1} – u_n \( \Rightarrow {v_n} = \frac{2}{{{3^n}}} \Rightarrow {v_2} = \frac{2}{{{3^2}}} = \frac{2}{9}\).

c) \(\mathop {\lim }\limits_{n \to + \infty } {v_n} = \mathop {\lim }\limits_{n \to + \infty } \frac{2}{{{3^n}}} = 0\).

d) 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 + … + v_{n – 1} = u_n – 2, ∀n ³ 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.