Ta có:\(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{10.10}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}< 1\)

=>A<1

\(\text{Ta có: }\frac{1}{2.2}< \frac{1}{1.2};\frac{1}{3.3}< \frac{1}{2.3};.....;\frac{1}{10.10}< \frac{1}{9.10}\)

 \(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{9.10}\)

\(\Rightarrow A< 1-\frac{1}{10}\)

\(\Rightarrow A< \frac{9}{10}< 1\)

A = 1/2.2 + 1/3.3 + 1/4.4 + ... + 1/10.10

A < 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/9.10

A < 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/9 - 1/10

A < 1 - 1/10 < 1

=> A < 1

Ta có:

\(\frac{1}{2.2}\)<\(\frac{1}{1.2}\)

\(\frac{1}{3.3}\)<\(\frac{1}{2.3}\)

..............

\(\frac{1}{1009.1009}\)<\(\frac{1}{1008.1009}\)

=>A< \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1008.1009}\)

=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1008}-\frac{1}{1009}\)

=\(\frac{1}{1}-\frac{1}{1009}=\frac{1008}{1009}>\frac{1008}{1344}=\frac{3}{4}\)

=>A<\(\frac{3}{4}\)

Mình nghĩ bạn cần xem lại :

\(A< \frac{1008}{1009}>\frac{1008}{1344}=\frac{3}{4}\)không có nghĩa là \(A< \frac{3}{4}\)

Xem lại ..

Đề bài sai rồi!Riêng 1/(2.2) đã bằng 1/4 rùi thì tổng trên phải lớn hơn 1/4 chứ!

Bạn Phạm Gia Bảo nói đúng đấy

Bạn nên sửa đề bài đi

Thấy :            \(\frac{1}{1.1!}=\frac{1}{1}\)

                       \(\frac{1}{2.2!}=\frac{1}{4}\)

                       \(\frac{1}{3.3!}< \frac{1}{1.2.3}\)( Vì 3.3! > 1.2.3 )

                         ...

                       \(\frac{1}{2019.2019!}< \frac{1}{2017.2018.2019}\)( vì 2019.2019! < 2017.2018.2019)

Cộng từng vế có :

  \(\frac{1}{3.3!}+\frac{1}{4.4!}+...+\frac{1}{2019.2019!}< \frac{1}{1.2.3}+...+\frac{1}{2017.2018.2019}\)

\(\Rightarrow\frac{1}{1.1!}+\frac{1}{2.2!}+...+\frac{1}{2019.2019!}< \frac{1}{1}+\frac{1}{4}+\frac{1}{1.2.3}+...+\frac{1}{2017.2018.2019}\)

\(\Rightarrow C< \frac{1}{1}+\frac{1}{4}+\left(\frac{1}{1.2}-\frac{1}{2.3}+...+\frac{1}{2017.2018}-\frac{1}{2018.2019}\right):2\)

\(\Rightarrow C< \frac{1}{1}+\frac{1}{4}+\left(\frac{1}{2}-\frac{1}{2018.2019}\right):2\)

\(\Rightarrow C< \frac{3}{2}-\frac{1}{2.2018.2019}\)

Vì \(\frac{1}{2.2018.2019}>0\Rightarrow C< \frac{3}{2}\)

Giúp mình bài này với

\(\frac{1}{2.2}+\frac{1}{4.4}+\frac{1}{6.6}+...+\frac{1}{200.200}\)

\(=\frac{1}{4}\left(1+\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\right)\)

\(< \frac{1}{4}\left(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

\(=\frac{1}{4}\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=\frac{1}{4}\left(1+1-\frac{1}{100}\right)=\frac{1}{4}\left(2-\frac{1}{100}\right)=\frac{1}{2}-\frac{1}{400}< \frac{1}{2}\)

Cảm ơn bạn

Ta có : \(\frac{1}{2.2}< \frac{1}{1.2}\)

            \(\frac{1}{3.3}< \frac{1}{2.3}\)

            \(\frac{1}{4.4}< \frac{1}{3.4}\)

              ...................

        \(\frac{1}{100.100}< \frac{1}{99.100}\)

Suy Ra : \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+......+\frac{1}{100.100}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{99.100}\)

\(\frac{1}{2.2}+\frac{1}{3.3}+.....+\frac{1}{100.100}< 1-\frac{1}{100}=\frac{99}{100}< 1\)

Ta có : \(\frac{1}{2.2}\)\(< \frac{1}{1.2}\)

                \(\frac{1}{3.3}\)\(< \frac{1}{2.3}\)

                 \(\frac{1}{4.4}\)\(< \frac{1}{3.4}\)

                   ......        ....   ......

              \(\frac{1}{100.100}\)\(< \frac{1}{99.100}\)

\(\Rightarrow\)\(\frac{1}{2.2}\)+ \(\frac{1}{3.3}\)+ \(\frac{1}{4.4}\)+ ..... + \(\frac{1}{100.100}\)< \(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ ..... + \(\frac{1}{99.100}\)

\(\frac{1}{2.2}\)+ \(\frac{1}{3.3}\)+ .... + \(\frac{1}{100.100}\)< \(1-\frac{1}{100}=\frac{99}{100}< 1\)

1/2.2 < 1/1.2

1/3.3 < 1/2.3

1/4.4 < 1/3.4

1/100.100   <   1/ 99.100

Nên 1/2.2 + 1/3.3 +1/4.4 + .... +1/100.100 < 1/1.1 +1/2.3+1/3.4 +......+ 1/99.100

1/2.2 + 1/3.3+.... 1/100.100 < 1 - 1/100 = 99/100 < 1

 ta còn có 1 cách làm ngắn gọn hơn