So sánh \(\frac{n}{n+1}\)và\(\frac{n-1}{n+4}\)(n\(\in\)N*)
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a) \({u_n} = \frac{{n + 1}}{n}= 1+ \frac{{1}}{n} > 1\).
b) \({u_n} = \frac{{n + 1}}{n}= 1+ \frac{{1}}{n} < 2\).
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 ~
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
\(\Rightarrow\frac{3}{n+3}>\frac{1}{n+2}\)
\(\Rightarrow1-\frac{3}{n+3}< 1-\frac{1}{n+2}\)
\(\Rightarrow\frac{n}{n+3}< \frac{n+1}{n+2}\)
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 \(\frac{n}{n+3}< \frac{n+1}{n+2}\)
a) Vì \(\frac{87}{39}>1\)
\(\frac{2015}{2017}< 1\)
\(\Rightarrow\frac{87}{39}>\frac{2015}{2017}\)
\(\frac{n}{n+1}\)và \(\frac{n+1}{n+3}\)
\(\Rightarrow\frac{n}{n+1}=\frac{n\cdot\left(n+3\right)}{\left(n+1\right)\left(n+3\right)}\)
\(\Rightarrow\frac{n+1}{n+3}=\frac{\left(n+1\right)^2}{\left(n+3\right)\left(n+1\right)}\)
\(\Rightarrow n\cdot\left(n+3\right)=n^2+3n\)
\(\Rightarrow\left(n+1\right)^2=n^2+2n+1\)
Dấu bằng chỉ xảy ra khi n = 1
Còn với mọi trường hợp n > 1 thì
\(\frac{n}{n+1}>\frac{n+1}{n+3};n^2+3n>n^2+2n+1\)
\(\frac{n}{n+1}=\frac{n\left(n+4\right)}{\left(n+1\right)\left(n+4\right)}>\frac{\left(n-1\right)\left(n+4\right)}{\left(n+1\right)\left(n+4\right)}>\frac{\left(n-1\right)\left(n+1\right)}{\left(n+1\right)\left(m+4\right)}=\frac{n-1}{n+4}\)