Sai phân CBH

Question

CMR:

$\frac{1}{2\sqrt{1}}$ +

$\frac{1}{3\sqrt{2}}$ +...+

$\frac{1}{2015\sqrt{2014}}<2$

score 1 · Answer 1

Tổng quát:

$\frac{1}{(n+1)\sqrt{n}}$ =

$\frac{\sqrt{n}}{(n+1)n}$

=

$\sqrt{n}. \frac{1}{n(n+1)}$

=

$\sqrt{n}.(\frac{1}{n} - \frac{1}{n+1})$

=

$\sqrt{n}.(\frac{1}{(\sqrt{n})^{2}}-\frac{1}{(\sqrt{n+1)}^{2}}$ )

=

$\sqrt{n}.(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}})(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}$ )

=

$(1+\frac{\sqrt{n}}{\sqrt{n+1}})(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}})$

(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 -

$\frac{2}{\sqrt{2015}}$ < 2

Vậy A < 2

1 Đáp án