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-A =( 1- 1/2 )(1 -1/3).....(1 -1/10)
= 1/2 . 2/3 ..... 9/10
= 1/10
-A = 1/10 nên A = -1/10
Vì 1/10 < 1/9 nên -1/10 > -1/9
Vậy A > -1/9
\(A=\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right)...\left(\frac{1}{10}-1\right)=-\frac{1}{2}.-\frac{2}{3}...-\frac{9}{10}\)
\(=\frac{-\left(1.2...9\right)}{2.3...10}=\frac{-1}{10}\)
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\)
\(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\)
Ta có: \(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\)\(< \)\(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\left(1\right)\)
Mà \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(=1-\frac{1}{n}< 1\left(2\right)\). Từ (1) và (2) suy ra
\(A< B< 1\Rightarrow A< 1\)
\(B=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)
\(B=-\frac{3}{2^2}.\left(-\frac{8}{3^2}\right).\left(-\frac{15}{4^2}\right)...\left(-\frac{9999}{100^2}\right)\)
\(B=-\left(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{9999}{100^2}\right)\)(Vì có 99 thừa số, mỗi thừa số là âm nên kết quả là âm)
\(B=-\left(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{99.101}{100.100}\right)\)
\(B=-\left(\frac{1.2.3...99}{2.3.4...100}.\frac{3.4.5...101}{2.3.4...100}\right)\)
\(B=-\left(\frac{1}{100}.\frac{101}{2}\right)\)
\(B=-\frac{101}{200}< -\frac{100}{200}=-\frac{1}{2}\)
\(\Rightarrow B< -\frac{1}{2}\)
Ta có :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(A< 1-\frac{1}{9}=\frac{8}{9}=B\)
\(\Rightarrow\)\(A< B\)
Vậy \(A< B\)
Chúc bạn học tốt ~
\(A=\left(\frac{1}{1^2}-1\right)\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2015^2}-1\right)\left(\frac{1}{2016^2}-1\right)\)
\(=0.\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2015^2}-1\right)\left(\frac{1}{2016^2}-1\right)=0>-\frac{1}{2}\)
suy ra A>B
\(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right)......\left(\frac{1}{10^2}-1\right)=\left(-\frac{3}{4}\right).\left(-\frac{8}{9}\right)......\left(-\frac{99}{100}\right)\)
\(A=\frac{\left(-3\right).\left(-8\right).....\left(-99\right)}{4.9........100}=\frac{\left(-1\right).3.\left(-2\right).4....\left(-9\right).11}{2.2.3.3.....10.10}=\frac{\left[\left(-1.-2.-3....-9\right).\left(3.4...11\right)\right]}{\left(2.3.....10\right).\left(2.3...10\right)}\)
\(A=\frac{\left(-1\right).11}{10.2}=\frac{-11}{20}< \frac{-10}{20}=\frac{-1}{2}\)
Suy ra \(A< -\frac{1}{2}\)