Chủ đề này có 1 trả lời

[TianaLoveEveryone]

1. Rút gọn:

 

$\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{(n-1)n(n+1)}$

 

Từ đó C/m:

$\frac{1}{3^{3}}+\frac{1}{4^{3}}+\frac{1}{5^{3}}+...+\frac{1}{(n+1)^{3}}<\frac{1}{12}$

 

2. CM:

 

a. $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2^{n}-1}> \frac{n}{2}$

 

b. $\frac{\sqrt{1}}{2}+\frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{3}+...+\frac{\sqrt{n}}{n}>\sqrt{n}$

Bài viết đã được chỉnh sửa nội dung bởi buiminhhieu: 12-06-2014 - 19:07

[BysLyl]

1. Rút gọn:

 

$\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{(n-1)n(n+1)}$

 

Từ đó C/m:

$\frac{1}{3^{3}}+\frac{1}{4^{3}}+\frac{1}{5^{3}}+...+\frac{1}{(n+1)^{3}}<\frac{1}{12}$

 

 

$\frac{1}{2}(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{(n-1)n(n+1)})=\frac{1}{2}(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{(n-1)n}-\frac{1}{n(n+1)})= \frac{1}{4}-\frac{1}{2n(n+1)}$

Xét $(k-1)(k+1)=k^{2}-1< k^{2}\Rightarrow (k-1)k(k+1)< k^{3} \Rightarrow \frac{1}{3^{3}}+\frac{1}{4^{3}}+...+\frac{1}{(n+1)^{3}}< \frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n(n+1)(n+2)}=\frac{1}{2.2.3}-\frac{1}{2(n+1)(n+2)}< \frac{1}{12}$