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24 tháng 4 2016

Với số tự nhiên \(n\ge2\) Ta có \(\frac{1}{\left(2n\right)^2}=\frac{1}{4}.\frac{1}{n^2}<\frac{1}{4}.\frac{1}{n\left(n-1\right)}\)Vậy \(B=\frac{1}{4}\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+.........+\frac{1}{n^2}\right)\)Và 
\(B<\frac{1}{4}\left(1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+................+\frac{1}{n-1}-\frac{1}{n}\right)\)Hay \(B<\frac{1}{4}\left(2-\frac{1}{n}\right)=\frac{1}{2}-\frac{1}{4n}<\frac{1}{2}\)
Vậy \(B<\frac{1}{2}\)

8 tháng 12 2015

\(A<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=1-\frac{1}{n}<1\)

 

20 tháng 9 2015

a) ta có: \(\frac{1}{2^2}

3 tháng 1 2018

Dễ vcl

2 tháng 12 2017

a) Ta có :

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\)

\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}=1-\frac{1}{n}< 1\)

\(\Rightarrow\)A < 1 

b) \(B=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)

\(B=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^3}+...+\frac{1}{n^2}\right)\)

vì \(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}< 2-\frac{1}{n}< 2\)

\(\Rightarrow B< \frac{1}{2^2}.2=\frac{1}{2}\)

2 tháng 12 2017

cảm ơn nha!

NV
2 tháng 4 2019

\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\)

\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)

\(A< 1-\frac{1}{n}< 1\)

Vậy \(A< 1\)

\(B=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)=\frac{1}{2^2}\left(1+A\right)\)

\(A< 1\Rightarrow B< \frac{1}{2^2}\left(1+1\right)=\frac{1}{2}\)

\(\Rightarrow B< \frac{1}{2}\)

10 tháng 10 2016

\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)

\(A=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\)

\(A< \frac{1}{2^2}.\left(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right).n}\right)\)

\(A< \frac{1}{4}.\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)

\(A< \frac{1}{4}.\left(2-\frac{1}{n}\right)\)

\(A< \frac{1}{4}.2=\frac{1}{2}\left(đpcm\right)\)

10 tháng 10 2016

\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)

\(\Rightarrow A=\frac{1}{\left(1.2\right)^2}+\frac{1}{\left(2.2\right)^2}+\frac{1}{\left(2.3\right)^2}+....+\frac{1}{\left(2n\right)^2}\)

\(\Rightarrow A=\frac{1}{1^2.2^2}+\frac{1}{2^2.2^2}+\frac{1}{2^2.3^2}+...+\frac{1}{2^2.n^2}\)

\(\Rightarrow A=\frac{1}{1}.\frac{1}{2^2}+\frac{1}{2^2}.\frac{1}{2^2}+\frac{1}{2^2}.\frac{1}{3^2}+...+\frac{1}{2^2}.\frac{1}{n^2}\)

\(\Rightarrow A=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2^2}+\frac{1}{n^2}\right)\)

Có: \(1+\frac{1}{2^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2^2}+\frac{1}{n^2}\) > 1

Rồi bạn tự tính tiếp nhé.