Bắng quy nạp, ta dễ chứng minh $\frac{a_{n}}{n}>1$ Ta tìm được $a_{n}=\sqrt[n+1]{\frac{n+1}{n}}+\sqrt[n]{\frac{n}{n-1}}+...\sqrt{\frac{2}{1}}$ Ta lại có: $\sqrt[k+1]{\frac{k+1}{k}}=\frac{\sqrt[k+1]{(k+1)k^{k}}}{k}<\frac{k+1+k^{2}}{(k+1)k}=1+\frac{ 1}{k}-\frac{1}{k+1}$. suy ra $a_{n}< n+1-\frac{1}{n+1}$. nên $\frac{a_{n}}{n}<1+\frac{1}{n}-\frac{1}{(n+1)n}$. Theo đl kẹp suy ra $lim\frac{a_{n}}{n}=1$ [RIGHT][I][B]Nguồn: MathScope.ORG[/B][/I][/RIGHT] __________________ chim chuột |