Tìm tổng \(S = \frac{1}{3} - \frac{1}{9} + \frac{1}{{27}} - ... + \frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{{3^n}}} + ...\).

Ta có \(1 + 2 + ... + n = \frac{{n\left( {n + 1} \right)}}{2}\).

Khi đó \(\mathop {\lim }\limits_{n \to + \infty } \frac{{1 + 2 + ... + n}}{{{n^2} + 3n}}\)\( = \mathop {\lim }\limits_{n \to + \infty } \frac{{n\left( {n + 1} \right)}}{{2\left( {{n^2} + 3n} \right)}}\)\( = \mathop {\lim }\limits_{n \to + \infty } \frac{{{n^2}\left( {1 + \frac{1}{n}} \right)}}{{2{n^2}\left( {1 + \frac{3}{n}} \right)}} = \frac{1}{2}\).

Trả lời: 0,5.

a) \(\mathop {\lim }\limits_{n \to + \infty } \frac{1}{{{v_n}}} = \mathop {\lim }\limits_{n \to + \infty } {\left( {\frac{1}{7}} \right)^n} = 0\).

b) \(\mathop {\lim }\limits_{n \to + \infty } {7^n} = + \infty \).

c) \(\mathop {\lim }\limits_{n \to + \infty } \frac{{{u_n} - {v_n}}}{{3{u_n} + 2{v_n}}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{{{4.3}^n} - {{8.7}^n}}}{{{{12.3}^n} - {{19.7}^n}}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{4.{{\left( {\frac{3}{7}} \right)}^n} - 8}}{{12.{{\left( {\frac{3}{7}} \right)}^n} - 19}} = \frac{8}{{19}}\).

d) Ta có \(\mathop {\lim }\limits_{n \to + \infty } {u_n} = \mathop {\lim }\limits_{n \to + \infty } \left( {{{4.3}^n} - {7^{n + 1}}} \right) = \mathop {\lim }\limits_{n \to + \infty } {7^n}\left[ {4.{{\left( {\frac{3}{7}} \right)}^n} - 7} \right]\).

Vì \(\mathop {\lim }\limits_{n \to + \infty } {7^n} = + \infty \); \(\mathop {\lim }\limits_{n \to + \infty } \left[ {4.{{\left( {\frac{3}{7}} \right)}^n} - 7} \right] = - 7 < 0\) nên \(\mathop {\lim }\limits_{n \to + \infty } {u_n} = - \infty \).

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