TD
0
K
Khách
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TD
5
8 tháng 12 2015
\(A<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=1-\frac{1}{n}<1\)
HT
0
ND
2
MT
16 tháng 7 2015
\(\text{a)}A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(=\frac{1}{1}-\frac{1}{n}=1-\frac{1}{n}<1\left(\text{vì n}\ge2\text{ hay n dương}\right)\)
Vậy A<1
K
6 tháng 8 2018
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VA
0
24 tháng 8 2019
Xét\(2C=2+1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2^{n-1}}\)
Mà\(C=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2^{n-1}}+\frac{1}{2^n}\)
\(\Rightarrow2C-C=2-\frac{1}{2^n}\Leftrightarrow C=2-\frac{1}{2^n}< 2\)
Vậy C<2
13 tháng 5 2019
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\)
\(A=\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+...+\frac{1}{n\cdot n}\)
\(A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{(n-1)\cdot n}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(A< 1-\frac{1}{n}\)
\(A< \frac{n-1}{n}< 1\)
13 tháng 5 2019
\(B=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{2n^2}\)
Theo câu a \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< 1\) nên \(B< \frac{1}{4}\cdot1=\frac{1}{4}\)