Tổng \(S = 1 + \frac{1}{3} + \frac{1}{{{3^2}}} + \frac{1}{{{3^3}}} + ... + \frac{1}{{{3^n}}} + ...\) có giá trị là

 

Ta có: \(\mathop {\lim }\limits_{n \to + \infty } \frac{{2022{n^2} - 2023n}}{{2024 + 2025n - 2026{n^2}}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{{n^2}\left( {2022 - \frac{{2023}}{n}} \right)}}{{{n^2}\left( {\frac{{2024}}{{{n^2}}} + \frac{{2025}}{n} - 2026} \right)}}\)

              \( = \mathop {\lim }\limits_{n \to + \infty } \frac{{2022 - \frac{{2023}}{n}}}{{\frac{{2024}}{{{n^2}}} + \frac{{2025}}{n} - 2026}}\)\( = \frac{{2022}}{{ - 2026}} = - \frac{{1011}}{{1013}}.\)

Chọn B

\(\lim \left( {{u_n} + 3} \right) = 0 \Leftrightarrow \lim {u_n} + \lim 3 = 0 \Leftrightarrow \lim {u_n} + 3 = 0 \Leftrightarrow \lim {u_n} = - 3.\)