TD
0
K
Khách
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HT
0
8 tháng 12 2015
\(A<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=1-\frac{1}{n}<1\)
YN
9 tháng 2 2023
Ta có:
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
...
\(\dfrac{1}{n^2}< \dfrac{1}{n\left(n-1\right)}\)
\(\Rightarrow P< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n-1\right)}\)
\(\Rightarrow P< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(\Rightarrow P< 1-\dfrac{1}{n}< 1\)
\(\Rightarrow P< 1\)
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
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}
K
6 tháng 8 2018
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VA
0