Ta có : 

\(\frac{1}{2^2}< \frac{1}{1.2}\)

\(\frac{1}{3^2}< \frac{1}{2.3}\)

\(\frac{1}{4^2}< \frac{1}{3.4}\)

\(............\)

\(\frac{1}{n^2}< \frac{1}{\left(n-1\right)n}\)

\(\Rightarrow\)\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\)

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

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

\(\Rightarrow\)\(A< 1-\frac{1}{n}< 1\)

Vậy \(A< 1\)

Chúc bạn học tốt ~ 

A đâu !!

anh cũng đang định hỏi câu này

Ta có \(A=\frac{1}{2!}+\frac{2}{3!}+...+\frac{2014}{2015!}\)

=>  \(A=\frac{2-1}{2!}+\frac{3-1}{3!}+...+\frac{2015-1}{2015!}\)

=>  \(A=\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+...+\frac{2015}{2015!}-\frac{1}{2015!}\)

=> \(A=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+...+\frac{1}{2014!}-\frac{1}{2015!}\)

=>  \(A=1-\frac{1}{2015!}< 1\)

Đặt biểu thức trung gian là :

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

Còn \(B=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{\left(n-1\right)\left(n+1\right)}\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{n-1}-\frac{1}{n+1}\right)\)

\(=\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\frac{1}{n+1}\right)\)

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

Vậy \(A< 3< \frac{3}{4}< 1.\)

Cách 2. Gọi biểu thức trên là A.Ta làm trội:

\(\frac{1}{x^2}\left(x\ge2\right)=\frac{1}{x.x}< \frac{1}{\left(x-1\right).x}\). Khi đó, áp dụng vào,ta có:

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

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

\(=1-\frac{1}{n}< 1\forall n\ge2^{\left(đpcm\right)}\)

a)\(\frac{1}{2^2}=\frac{1}{2.2}<\frac{1}{1.2};\frac{1}{3^2}=\frac{1}{3.3}<\frac{1}{2.3};...;\frac{1}{n^2}=\frac{1}{n.n}<\frac{1}{\left(n-1\right).n}\)

=>\(A<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{\left(n-1\right).n}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{n-1}-\frac{1}{n}\)

=>A<1-1/n

mà 1-1/n<1

=>A<1

b)tương tự

a)Ta có: A= \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{n^2}\Rightarrow A<\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{\left(n-1\right)n}\)

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

\(\Rightarrow A<1-\frac{1}{n}\Rightarrow A<1\)

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

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

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

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

Đặt \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)\cdot n}\)

Ta có:

\(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\)\(< \)\(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)\cdot n}\left(1\right)\) 

Mà \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)\cdot n}\)

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

\(=1-\frac{1}{n}< 1\left(2\right)\)(đúng. vì \(n\ge2\))

Từ (1) và (2) \(\Rightarrow A< B< 1\Rightarrow A< 1\)

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