a: \(A=\frac12+\frac{1}{2^2}+\cdots+\frac{1}{2^{2022}}\)
=>\(2A=1+\frac12+\cdots+\frac{1}{2^{2021}}\)
=>2A-A=\(1+\frac12+\cdots+\frac{1}{2^{2021}}-\frac12-\frac{1}{2^2}-\cdots-\frac{1}{2^{2022}}\)
=>\(A=1-\frac{1}{2^{2022}}\)
=>A<1
b: Ta có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)
\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)
...
\(\frac{1}{n^2}<\frac{1}{\left(n-1\right)\cdot n}=\frac{1}{n-1}-\frac{1}{n}\)
Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}<1-\frac12+\frac12-\frac13+\cdots+\frac{1}{n-1}-\frac{1}{n}\)
=>\(M<1-\frac{1}{n}\)
=>M<1
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