chung to: 1/22 + 1/32+1/42+....+1/19632 <1
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1 tháng 5 2016
\(\frac{1}{8}=\frac{1}{8}\)
\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}<\frac{3}{10}\)
\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}<\frac{3}{40}\)
-> A <\(\frac{1}{8}+\frac{3}{10}+\frac{3}{40}=\frac{20}{40}=\frac{1}{2}\)
22 tháng 6 2023
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
Ta có: \(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};...;\frac{1}{1963^2}<\frac{1}{1962.1963}\)
Cộng vế theo vế ta được: \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1963^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1962.1963}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1962.1963}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1962}-\frac{1}{1963}\)
\(=\)\(1-\frac{1}{1963}<1\)
Do đó :\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1963^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1962.1963}\)\(<1\)