a) \({u_n} = \frac{{n + 1}}{n}= 1+ \frac{{1}}{n} > 1\).

b) \({u_n} = \frac{{n + 1}}{n}= 1+ \frac{{1}}{n} < 2\).

n/n+3=n:(n+3)=n:n+n:3=1+n:3

n+1/n+2=(n+1):(n+2)=(n+1):n+(n+1):(n+2)=1+n+n/2+1/2=3/2+3n/2=3(1+n):2

Vì ta thấy rõ 3(1+n):2 > 1+n :3 

Hay n/n+3 < n+1/n+2

Ta xét 2 phân số sau thì có :

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

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

Để so sánh 2 phân số trên ta so sánh\(\frac{3}{n+3};\frac{1}{n+2}\)

Quy đồng lên ta có :

\(\frac{3}{n+3}=\frac{3\left(n+2\right)}{\left(n+3\right)\left(n+2\right)}=\frac{3n+6}{\left(n+3\right)\left(n+2\right)}\)

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

Mà 3n+6>n+3

\(\Rightarrow\frac{3}{n+3}>\frac{1}{n+2}\)

\(\Rightarrow1-\frac{3}{n+3}< 1-\frac{1}{n+2}\)

\(\Rightarrow\frac{n}{n+3}< \frac{n+1}{n+2}\)

Mọi người làm vại chi cho phức tạp.

Ta có: \(\frac{n}{n+3}< \frac{n}{n+2}\) (vì \(n+3>n+2\))

Và \(\frac{n}{n+2}< \frac{n+1}{n+2}\)(vì \(n< n+1\))

Theo tính chất bắc cầu suy ra \(\frac{n}{n+3}< \frac{n+1}{n+2}\)

Ta có : 

\(\frac{n}{n+3}< \frac{n}{n+2}\)

\(\frac{n+1}{n+2}>\frac{n}{n+2}\)

\(\Rightarrow\frac{n}{n+3}< \frac{n}{n+2}< \frac{n+1}{n+2}\)

Vậy \(\frac{n}{n+3}< \frac{n+1}{n+2}\)

Cách 1 :

Ta có : \(\frac{n}{n+1}>\frac{n}{2n+3}\left(1\right)\)

          \(\frac{n+1}{n+2}>\frac{n+1}{2n+3}\left(2\right)\)

Cộng theo từng vế ( 1) và ( 2 ) ta được :

\(A=\frac{n}{n+1}+\frac{n+1}{n+2}>\frac{2n+1}{2n+3}=B\)

VẬY \(A>B\)

CÁCH 2

\(A=\frac{n}{n+1}+\frac{n+1}{n+2}>\frac{n}{n+2}+\frac{n+1}{n+2}\)

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

VẬY A>B  

Chúc bạn học tốt ( -_- )

\(\frac{n+1}{n+2}\)và \(\frac{n}{n+3}\)

<=>\(\hept{\begin{cases}\left(n+1\right).\left(n+3\right)=n^2+4n+3\\\left(n+2\right).n=n^2+2n\end{cases}}\)

<=>\(n^2\)+4n+3 > \(n^2\)+2n

<=>\(\left(n+1\right).\left(n+3\right)>\left(n+2\right).n\)

<=>\(\frac{n+1}{n+2}>\frac{n}{n+3}\)

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 ~