Question

Chứng minh bất đẳng thức sau:

a/ \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{\left(2n-1\right).\left(2n+1\right)}<\frac{1}{2}\)

b/ \(1+\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3...n}<2\)

Giúp mình với!!!!!!!!!!!!!!!!

Answer 1

bạn làm theo công thức \(\frac{n}{n.\left(n+1\right)}=\frac{n}{n}-\frac{n}{n+1}\)

Answer 2

a)Đặt A= \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{\left(2n-1\right)\left(2n+1\right)}\)

\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}\)

\(\Rightarrow2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n-1}-\frac{1}{2n+1}\)

\(\Rightarrow2A=1-\frac{1}{2n+1}< 1\)

\(\Rightarrow A< \frac{1}{2}\)(đpcm)

b)Ta có: \(1+\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3...n}< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\)

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

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

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

\(\Rightarrow1+\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3...n}< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}< 2\)

\(\Rightarrow1+\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3...n}< 2\)(đpcm)

Answer 3

\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2n-1\right).\left(2n+1\right)}=2A\)

\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n-1}-\frac{1}{2n+1}=2A\)

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

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

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

Mà

\(\frac{1}{2}=\frac{1.n}{2.n}=\frac{n}{2n}\)

vì 2A  =\(\frac{2n}{2n+1}\)

suy ra A = \(\frac{1n}{2n+1}\)

vì mẫu của \(\frac{1}{2}\)bé hơn mẫu của A  

suy ra A < \(\frac{1}{2}\)

suy ra ĐPCM