\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\)

\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)

\(=1-\frac{1}{n}< 1\)( vì n \(\ge\)2 )

a) ta có: \(\frac{1}{2^2}

Dễ vcl

Cho n $\in$∈ N và n $\ge$≥ 2. Hãy so sánh.A= $\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+............+\frac{1}{n^2}$122 +132 +142 +............+1n2 với 1 tịk nhé cho tròn 160

với k>=2:

1/k² < 1/k(k-1) = (k-(k-1))/k(k-1) =1/(k-1) +1/k

apf dụng với k=2,3,...,n sẽ tính được A<1

a) Ta có :

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\)

\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}=1-\frac{1}{n}< 1\)

\(\Rightarrow\)A < 1 

b) \(B=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)

\(B=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^3}+...+\frac{1}{n^2}\right)\)

vì \(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}< 2-\frac{1}{n}< 2\)

\(\Rightarrow B< \frac{1}{2^2}.2=\frac{1}{2}\)

cảm ơn nha!

\(A<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=1-\frac{1}{n}<1\)

Với số tự nhiên \(n\ge2\) Ta có \(\frac{1}{\left(2n\right)^2}=\frac{1}{4}.\frac{1}{n^2}<\frac{1}{4}.\frac{1}{n\left(n-1\right)}\)Vậy \(B=\frac{1}{4}\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+.........+\frac{1}{n^2}\right)\)Và 
\(B<\frac{1}{4}\left(1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+................+\frac{1}{n-1}-\frac{1}{n}\right)\)Hay \(B<\frac{1}{4}\left(2-\frac{1}{n}\right)=\frac{1}{2}-\frac{1}{4n}<\frac{1}{2}\)
Vậy \(B<\frac{1}{2}\)