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Ta có:
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
...
\(\dfrac{1}{n^2}< \dfrac{1}{n\left(n-1\right)}\)
\(\Rightarrow P< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n-1\right)}\)
\(\Rightarrow P< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(\Rightarrow P< 1-\dfrac{1}{n}< 1\)
\(\Rightarrow P< 1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\text{a)}A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{n\left(n-1\right)}\\ A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}=1-\dfrac{1}{n}< 1\left(\dfrac{1}{n}>0\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi ƯCLN(2n+1;2n^2-1)=d
Ta có: 2n+1 chia hết cho d; 2n2-1 chia hết cho d
=>n(2n+1) chia hết cho d; 2n^2-1 chia hết cho d
=>2n^2+2 chia hết cho d; 2n^2-1 chia hết cho d
=>2n^2+2-2n^2-1 chia hết cho d
hay 1 chia hết cho d hay d=1
nên ƯCLN(2n+1;2n^2-1)=1
Vậy A là ps tối giản với mọi n
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=1-\frac{1}{n}<1\)
Do
với mọi n ≥ 2 nên
A < C =![\frac{1}{{{2^2} - 1}} + \frac{1}{{{3^2} - 1}} + ... + \frac{1}{{{n^2} - 1}}](https://tex.vndoc.com/?tex=%5Cfrac%7B1%7D%7B%7B%7B2%5E2%7D%20-%201%7D%7D%20%2B%20%5Cfrac%7B1%7D%7B%7B%7B3%5E2%7D%20-%201%7D%7D%20%2B%20...%20%2B%20%5Cfrac%7B1%7D%7B%7B%7Bn%5E2%7D%20-%201%7D%7D)
Mặt khác:
Vậy A < 1
b.
\(\Rightarrow P< 0,5\)