Bài 7 (1.0 điểm). Cho dãy số $\left(u_n\right)$ biết $\left\{\begin{array}{l}u_1=1 \\ u_{n+1}=u_n+\frac{u_n^2}{2023}\end{array}, \forall n \in \mathbb{N}^*\right.$. Tính $\lim \left(\frac{u_1}{u_2}+\frac{u_2}{u_3}+\ldots+\frac{u_n}{u_{n+1}}\right)$.

Ta có:

`u_{n+1} = u_n + \frac{u_n^2}{2023}`

`\frac{u_{n+1}}{u_n} = 1 + \frac{u_n}{2023}`

`\frac{u_n}{u_{n+1}} = \frac{1}{1 + \frac{u_n}{2023}}`

Xét `\frac{1}{u_{n+1}} = \frac{1}{u_n(1 + \frac{u_n}{2023})} = \frac{1}{u_n} - \frac{1}{2023 + u_n}`

Suy ra: `\frac{1}{u_n} - \frac{1}{u_{n+1}} = \frac{1}{2023 + u_n}`

Do đó: `\frac{u_n}{u_{n+1}} = \frac{1}{1 + \frac{u_n}{2023}} = \frac{2023}{2023 + u_n}`

Ta có:

`\frac{1}{u_n} - \frac{1}{u_{n+1}} = \frac{1}{2023 + u_n}`

`\sum_{i=1}^{n} (\frac{1}{u_i} - \frac{1}{u_{i+1}}) = \sum_{i=1}^{n} \frac{1}{2023 + u_i}`

`\frac{1}{u_1} - \frac{1}{u_{n+1}} = \sum_{i=1}^{n} \frac{1}{2023 + u_i}`

`\frac{1}{u_{n+1}} = 1 - \sum_{i=1}^{n} \frac{1}{2023 + u_i}`

Vì `u_1 = 1` ta có `u_n` là dãy số dương và tăng

`\frac{u_{n+1}}{u_n} = 1 + \frac{u_n}{2023} > 1`

Suy ra `u_n` là dãy tăng

Mặt khác `\frac{1}{u_{n+1}} = \frac{1}{u_n} - \frac{u_n}{2023u_n + u_n^2} < \frac{1}{u_n}`

Dãy `\frac{1}{u_n}` giảm. Do đó tồn tại giới hạn hữu hạn của `\frac{1}{u_n}`

Đặt `\lim \frac{1}{u_n} = a`

Suy ra `\lim \frac{u_n}{2023} = 0`

`\lim \frac{u_{n+1}}{u_n} = 1`

`\lim (\frac{u_1}{u_2} + \frac{u_2}{u_3} + ... + \frac{u_n}{u_{n+1}}) = \lim \sum_{i=1}^{n} \frac{u_i}{u_{i+1}} = \lim \sum_{i=1}^{n} \frac{2023}{2023 + u_i}`

`\lim \sum_{i=1}^{n} (\frac{1}{u_i} - \frac{1}{u_{i+1}}) = \frac{1}{u_1} = 1`

`\lim \frac{1}{u_{n+1}} = 0`

`\lim \sum_{i=1}^{n} \frac{1}{2023 + u_i} = 1`

Vậy `\lim \sum_{i=1}^{n} \frac{u_i}{u_{i+1}} = 2023`