với mọi số tự nhiên n lớn hơn hoặc bằng 2 hãy so sánh :
a/ A = \(\frac{1}{2^2}\)+ \(\frac{1}{3^2}\)+ \(\frac{1}{4^2}\)+.....+\(\frac{1}{n^2}\)với 1
b/ B = \(\frac{1}{2^2}\)+ \(\frac{1}{4^2}\)+ \(\frac{1}{6^2}\)+.....+\(\frac{1}{\left(2n\right)^2}\)với \(\frac{1}{2}\)
nhanh nha mí bạn ......nhanh và đúng nhất mik sẽ tick cho nha ^^
a,Xét thấy: \(A\le\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{\left(n-1\right)n}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}=1-\frac{1}{n}\le1\)
b, Xét thấy : \(B=\frac{1}{4}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)\)
\(\le\frac{1}{4}\left[1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{\left(n-1\right)n}\right]\)
\(=\frac{1}{4}\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)=\frac{1}{4}\left(2-\frac{1}{n}\right)=\frac{1}{2}-\frac{1}{4n}\le\frac{1}{2}\)
