$\mathop{\lim }\limits_{n\to\infty }$2013$\left(1+\frac{2}{4025^{2^{n-1}}-1}\right)$
Chú ý rằng $\mathop {\lim }\limits 2^{n-1}= +\infty  $ do đó $\mathop {\lim }\limits_{n \to \infty } \frac{1}{{{{4025}^{{2^{n - 1}}}} - 1}} = 0$
Suy ra đáp số cần tìm là $I=2003.$
à, xin lỗi bạn, có thể cho mình hỏi " Chú ý rằng lim2^(n−1)= ∞ là kiến thức trong sách giáo khoa phải không, do mình chỉ mới học tới bài Giới hạn của dãy số, trong phần công thức không có đề cập dến điều bạn vừa nêu trên, kg biet mình có ap dung dk kg ? –  thattiennu_kute_dangiu 31-05-13 11:17 PM
Do $\mathop {\lim }\limits_{n \to \infty } \frac{2}{{{{4025}^{{2^{n - 1}}}} - 1}} = 0$ suy ra $\mathop {\lim }\limits_{n \to \infty } 2013\left( {1 + \frac{2}{{{{4025}^{{2^{n - 1}}}} - 1}}} \right) = 2013$.

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