So sánh:
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{n^2}\)
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8 tháng 12 2015
\(A<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=1-\frac{1}{n}<1\)
24 tháng 6 2018
Ta có :
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(............\)
\(\frac{1}{n^2}< \frac{1}{\left(n-1\right)n}\)
\(\Rightarrow\)\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\)
\(\Rightarrow\)\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\)
\(\Rightarrow\)\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(\Rightarrow\)\(A< 1-\frac{1}{n}< 1\)
Vậy \(A< 1\)
Chúc bạn học tốt ~
15 tháng 1 2019
Sửa đề : \(A=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+......+\frac{1}{2^{199}}\)
\(\Rightarrow2A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+......+\frac{1}{2^{198}}\)
\(\Rightarrow2A-A=A=\frac{1}{2}-\frac{1}{2^{199}}< \frac{1}{2}+\frac{1}{4}=\frac{3}{4}\)
Vậy \(A< \frac{3}{4}\)
4 tháng 1 2018
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\)
\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(=1-\frac{1}{n}< 1\)( vì n \(\ge\)2 )
GJ
2
2 tháng 2 2019
Ta thấy rằng: \(2^2>1\times2\) , \(3^2>2\times3\),..., \(2011^2>2010\times2011\).
\(\Rightarrow A< \frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{2010\times2011}=\frac{2-1}{1\times2}+\frac{3-2}{2\times3}+...+\frac{2011-2010}{2010\times2011}\)\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2010}-\frac{1}{2011}\)\(=1-\frac{1}{2011}< 1.\)
Vậy A < 1.
9 tháng 10 2016
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) \(\Rightarrow A< B< 1\Rightarrow A< 1\)
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 \(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}< 2-\frac{1}{n}< 2\)
\(\Rightarrow B< \frac{1}{2^2}.2=\frac{1}{2}\)
\(\frac{1}{2^2}< \frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}\right)\)
\(\frac{1}{3^2}< \frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}\right)\)
\(\frac{1}{4^2}< \frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}\right)\)
\(\frac{1}{5^2}< \frac{1}{2}\left(\frac{1}{4}-\frac{1}{6}\right)\)
\(\frac{1}{\left(n-1\right)^2}< \frac{1}{2}\left(\frac{1}{n-2}-\frac{1}{n}\right)\)
\(\frac{1}{n^2}< \frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)\)
\(A< 1\)