Tổng quát: \frac{1}{(n+1)\sqrt{n}} = \frac{\sqrt{n}}{n(n+1)} = \sqrt{n}\frac{1}{n(n+1)} = \frac{n}{\frac{1}{n} - \frac{1}{n+1}} = \sqrt{n}(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}})(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}) =
Tổng quát:
$ \frac{1}{(n+1)\sqrt{n}}
$ =
$\frac{\sqrt{n}}{(n+1)
n}
$ =
$\sqrt{n}
. \frac{1}{n(n+1)}
$ =$\sqrt{n}.(\frac{1}{n} - \frac{1}{n+1})$ =
$\
sqr
t{n}
.(\frac{1}{
(\sqrt{n}
)^{2}}-\frac{1}{
(\sqrt{n+1
)}
^{2}
}$) =
$\sqrt{n}
.(\frac{1}{\sqrt{n}}
+\frac{1}{\sqrt{n+1}})(\frac{1}{\sqrt{n}}
-\frac{1}{\sqrt{n+1}}
$)
= $(1+\frac{\sqrt{n}}{\sqrt{n+1}})(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}})$ (Do $\frac{\sqrt{n}}{\sqrt{n+1}} = \sqrt{\frac{n}{n+1}} < \sqrt{\frac{n+1}{n+1}} = \sqrt{1} = 1$) < (1+1)($\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}})$ < 2($\frac{1}{\sqrt{n}}- \frac{1}{\sqrt{n+1}})$Cho n=1;2;3;...;2014. Ta có: (Gọi A là biểu thức cần CM)A < 2 (1 - $\frac{1}{\sqrt{2015}}$) = 2 - $\frac{2}{\sqrt{2015}}$ < 2Vậy A < 2