\(\text { Tính } \lim \frac{u_{n}}{n} \text { với } u_{n}=\sqrt{1+\frac{1}{1^{2}}+\frac{1}{2^{2}}}+\sqrt{1+\frac{1}{2^{2}}+\frac{1}{3^{2}}}+\ldots+\sqrt{1+\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}} \text { . }\)
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Lời giải:
Báo sai\(\begin{aligned} &\text { Ta có } \sqrt{1+\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}}=\sqrt{\frac{n^{2}(n+1)^{2}+(n+1)^{2}+n^{2}}{n^{2}(n+1)^{2}}} \\ &=\sqrt{\frac{n^{2}\left(n^{2}+2 n+1+1\right)+(n+1)^{2}}{n^{2}(n+1)^{2}}}=\sqrt{\frac{n^{4}+2 n^{2}(n+1)+(n+1)^{2}}{n^{2}(n+1)^{2}}} \\ &=\sqrt{\frac{\left(n^{2}+n+1\right)^{2}}{n^{2}(n+1)^{2}}}=\frac{n^{2}+n+1}{n(n+1)}=1+\frac{1}{n(n+1)}=1+\frac{1}{n}-\frac{1}{n+1} . \\ &\text { Suy ra } u_{n}=\sum_{k=1}^{n} \sqrt{1+\frac{1}{k^{2}}+\frac{1}{(k+1)^{2}}}=\sum_{k=1}^{n}\left(1+\frac{1}{k}-\frac{1}{k+1}\right)=n+1-\frac{1}{n+1} . \\ &\text { Vậy } \lim \frac{u_{n}}{n}=\lim \frac{n+1-\frac{1}{n+1}}{n}=\lim \frac{n^{2}+2 n}{n(n+1)}=\lim \frac{1+\frac{2}{n}}{1+\frac{1}{n}}=1 \end{aligned}\)
