\(\text { Tính } \lim u_{n} \text { với } u_{n}=\frac{n+\frac{n-1}{2}+\frac{n-2}{3}+\ldots+\frac{1}{n}}{\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n+1}}\)
Suy nghĩ và trả lời câu hỏi trước khi xem đáp án
Lời giải:
Báo sai\(\begin{aligned} &\text { Ta có } u_{n}=\frac{n+\left(\frac{n-1}{2}+\frac{1}{2}\right)+\left(\frac{n-2}{3}+\frac{2}{3}\right)+\ldots+\left(\frac{1}{n}+\frac{n-1}{n}\right)+\frac{n}{n+1}-\frac{1}{2}-\frac{2}{3}-\ldots-\frac{n}{n+1}}{\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n+1}} \\ &=\frac{\frac{n}{2}+\frac{n}{3}+\ldots+\frac{n}{n}+\frac{n}{n+1}+\left(n-\frac{1}{2}-\frac{2}{3}-\ldots-\frac{n}{n+1}\right)}{\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n+1}} \\ &=\frac{n\left(\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n+1}\right)+1-\frac{1}{2}+1-\frac{2}{3}+\ldots+1-\frac{n}{n+1}}{\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n+1}} \end{aligned}\)
\(\begin{aligned} &=\frac{n\left(\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n+1}\right)+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n+1}}{\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n+1}}=n+1 \\ &\text { Vậy } \lim u_{n}=\lim (n+1)=+\infty \end{aligned}\)