Ta có :
[TEX]u_{n+1}=(1+\frac{1}{n+1})^{n+2}[/TEX]
[TEX]\lim_{x\to 0} (1+\frac{1}{n})^{n+1}=\lim_{x\to 0} (1+\frac{1}{n})^n .(1+\frac{1}{n})^{n.\frac{1}{n}}=e.e^{\frac{1}{n}}[/TEX]
[TEX]\lim_{x\to 0} (1+\frac{1}{n+1})^{n+2}=\lim_{x\to 0} (1+\frac{1}{n+1})^{n+1} .(1+\frac{1}{n+1})^{\frac{n+1}{n+1}}=e.e^{\frac{1}{n+1}}[/TEX]
Do [TEX]\frac{1}{n+1} < \frac{1}{n} -->>[/TEX] dãy số giảm b-(