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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 \(B<\frac{1}{2}\)
\(A<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=1-\frac{1}{n}<1\)
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 \(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}< 2-\frac{1}{n}< 2\)
\(\Rightarrow B< \frac{1}{2^2}.2=\frac{1}{2}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(A< 1-\frac{1}{n}< 1\)
Vậy \(A< 1\)
\(B=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)=\frac{1}{2^2}\left(1+A\right)\)
Mà \(A< 1\Rightarrow B< \frac{1}{2^2}\left(1+1\right)=\frac{1}{2}\)
\(\Rightarrow B< \frac{1}{2}\)
\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)
\(A=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\)
\(A< \frac{1}{2^2}.\left(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right).n}\right)\)
\(A< \frac{1}{4}.\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)
\(A< \frac{1}{4}.\left(2-\frac{1}{n}\right)\)
\(A< \frac{1}{4}.2=\frac{1}{2}\left(đpcm\right)\)
\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)
\(\Rightarrow A=\frac{1}{\left(1.2\right)^2}+\frac{1}{\left(2.2\right)^2}+\frac{1}{\left(2.3\right)^2}+....+\frac{1}{\left(2n\right)^2}\)
\(\Rightarrow A=\frac{1}{1^2.2^2}+\frac{1}{2^2.2^2}+\frac{1}{2^2.3^2}+...+\frac{1}{2^2.n^2}\)
\(\Rightarrow A=\frac{1}{1}.\frac{1}{2^2}+\frac{1}{2^2}.\frac{1}{2^2}+\frac{1}{2^2}.\frac{1}{3^2}+...+\frac{1}{2^2}.\frac{1}{n^2}\)
\(\Rightarrow A=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2^2}+\frac{1}{n^2}\right)\)
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.
