CMR với mọi số tự nhiên n\(\ge\) 2, ta có : \(\frac{1}{2^3}+\frac{1}{3^3}+...+\frac{1}{n^3}< \frac{1}{4}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...+\frac{1}{n^3}
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có \(\frac{1}{n+k}>\frac{1}{2n}\)với k=1;2;...;n-1
=> \(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}>\frac{1}{2n}+\frac{1}{2n}+\frac{1}{2n}+....+\frac{1}{2n}=\frac{n}{2n}=\frac{1}{2}\)
Mặt khác ta có \(\frac{1}{n+k}+\frac{1}{n\left(+\left(n+1-k\right)\right)}< \frac{3}{2n}\)
\(\Leftrightarrow3k^2+3nk+n+3k\forall k=1;2;...;n\)
Với k=1 ta có \(\frac{1}{n+1}+\frac{1}{n+n}< \frac{3}{2n}\)
Với k=2 ta có \(\frac{1}{n+2}+\frac{1}{n+\left(n-1\right)}< \frac{3}{2n}\)
..........................................
Với k=n ta có \(\frac{1}{n+n}+\frac{1}{n+1}< \frac{3}{2n}\)
Cộng từng vế của 2 BĐT trên ta được
\(2\left(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}\right)< \frac{3}{2n}+\frac{3}{2n}+....+\frac{3}{2n}=\frac{3n}{2n}=\frac{3}{2}\)
\(\Rightarrow\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}< \frac{3}{4}\)(đpcm)
Không cần chứng minh \(\frac{1}{2}< \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
vì bài dài quá nên mình làm từng bài 1 nhé
1. Ta thấy : \(\frac{1}{n^3}< \frac{1}{n^3-n}=\frac{1}{\left(n-1\right)n\left(n+1\right)}=\frac{1}{2}.\frac{\left(n+1\right)-\left(n-1\right)}{\left(n-1\right)n\left(n+1\right)}=\frac{1}{2}.\left[\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right]\)
Do đó :
\(B< \frac{1}{2}.\left[\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right]< \frac{1}{2}.\frac{1}{6}=\frac{1}{12}\)
2.
Nhận xét : \(1+\frac{1}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)
Do đó :
\(A=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{\left(n+1\right)^2}{n\left(n+2\right)}=\frac{2.3...\left(n+1\right)}{1.2...n}.\frac{2.3...\left(n+1\right)}{3.4...\left(n+2\right)}=\frac{n+1}{1}.\frac{2}{n+2}< 2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)\cdot n}\)
Ta có:
\(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\)\(< \)\(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)\cdot n}\left(1\right)\)
Mà \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)\cdot n}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(=1-\frac{1}{n}< 1\left(2\right)\)(đúng. vì \(n\ge2\))
Từ (1) và (2) \(\Rightarrow A< B< 1\Rightarrow A< 1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\)
\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(=1-\frac{1}{n}< 1\)( vì n \(\ge\)2 )
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{3^3}< \frac{1}{2.3.4}\) \(\frac{1}{4^3}< \frac{1}{3.4.5}\) \(\frac{1}{5^3}< \frac{1}{4.5.6}\) ..... \(\frac{1}{n^3}< \frac{1}{\left(n-1\right)n\left(n+1\right)}\)
\(\Rightarrow B< \frac{1}{2.3.4}+\frac{1}{3.4.5}+\frac{1}{4.5.6}+...+\frac{1}{\left(n-1\right)n\left(n+1\right)}\)
\(\Rightarrow B< \frac{1}{2}\left(\frac{2}{2.3.4}+\frac{2}{3.4.5}+\frac{2}{4.5.6}+...+\frac{2}{\left(n-1\right)n\left(n+1\right)}\right)\)
\(\Rightarrow B< \frac{1}{2}\left(\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+\frac{6-4}{4.5.6}+...+\frac{\left(n+1\right)-\left(n-1\right)}{\left(n-1\right)n\left(n+1\right)}\right)\)
\(\Rightarrow B< \frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{4.5}-\frac{1}{5.6}+...+\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right)\)
\(\Rightarrow B< \frac{1}{2}\left(\frac{1}{6}-\frac{1}{n\left(n+1\right)}\right)=\frac{1}{12}-\frac{1}{2n\left(n+1\right)}< \frac{1}{12}\)
gọi A là vế trái của bất đẳng thức trên
Ta có : \(\frac{1}{k^3}< \frac{1}{k^3-k}=\frac{1}{k.\left(k-1\right)\left(k+1\right)}\)
Do đó : A < \(\frac{1}{2^3-2}+\frac{1}{3^3-3}+...+\frac{1}{n^3-n}=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{\left(n-1\right)n\left(n+1\right)}\)
Đặt C = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{\left(n-1\right)n\left(n+1\right)}\)
Ta thấy \(\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}=\frac{2}{\left(n-1\right)n\left(n+1\right)}\)
nên
C = \(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{n\left(n+1\right)}\right)=\frac{1}{4}-\frac{1}{2n\left(n+1\right)}< \frac{1}{4}\)
Vậy ....