\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=\left(1+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right).\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\right)\)\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)

\(A=\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{35}\right)+\left(\frac{1}{36}+...+\frac{1}{50}\right)>\frac{1}{35}.10+\frac{1}{50}.15=\frac{41}{70}>\frac{7}{12}\)

\(A< \frac{10}{26}+\frac{15}{36}< \frac{5}{6}\) Vậy ....

* Ta có : 1/21 >1/30 ;1/22 >1/30 ;...;1/29 >1/30 

=> 1/21 +1/22 +...+1/29 +1/30 >1/30 +1/30 +...+1/30 =10/30 =1/3    (1)

1/31 >1/40 ;1/32 >1/40 ;...;1/39 >1/40 

=> 1/31 +1/32 +...+1/39 +1/30 >1/40 +1/40 +...+1/40 =10/40 =1/4    (2)

Từ (1) và (2) 

=> 1/21 +1/22 +...+1/30 +1/31 +1/32 +...+1/40 >1/3 +1/4 

=> 1/21 +1/22 +1/23 +...+1/40 >7/12   (*)

* Ta có : 1/21 <1/20 ;1/22 <1/20 ;...;1/30 <1/20 

=> 1/21 +1/22 +...+1/29 +1/30 <1/20 +1/20 +...+1/20 =10/20 =1/2   (3)

1/31 <1/30 ;1/32 <1/30 ;...;1/40 <1/30 

=> 1/31 +1/32 +...+1/39 +1/40 <1/30 +1/30 +...+1/30 =10/30 =1/3   (4)

Từ (3) và (4) 

=> 1/21 +1/22 +...+1/30 +1/31 +1/32 +...+1/40 <1/2 +1/3 

=> 1/21 +1/22 +1/23+...+1/40 <5/6     (**)

Từ (*) và (**) ta có : 7/12 <1/21 +1/22 +1/23 +...+1/40 <5/6   (đpcm)

Bài hơi dài , thông cảm

Ta có : \(\frac{1}{21}>\frac{1}{30};\frac{1}{22}>\frac{1}{30};\frac{1}{23}>\frac{1}{30};...;\frac{1}{29}>\frac{1}{30}\)

\(\Rightarrow A=\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{29}>\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\)

\(>\frac{10}{30}=\frac{1}{3}(1)\)

Ta có  : \(\frac{1}{31}>\frac{1}{40},\frac{1}{32}>\frac{1}{40},...,\frac{1}{39}>\frac{1}{40}\)

\(\Rightarrow A=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{39}>\frac{1}{40}+\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\)

\(>\frac{10}{40}=\frac{1}{4}(2)\)

Từ 1 và 2 \(\Rightarrow A>\frac{1}{3}+\frac{1}{4}\Rightarrow A>\frac{7}{12}\)

Ta có : \(\frac{1}{21}< \frac{1}{20};\frac{1}{22}< \frac{1}{20};...;\frac{1}{30}< \frac{1}{20}\)

\(\Rightarrow A=\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{30}< \frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\)

\(< \frac{10}{20}=\frac{1}{2}(3)\)

Ta lại có : ....

Làm tiếp đi :v

Thôi,làm nốt :v

Ta lại có : \(\frac{1}{31}< \frac{1}{30};\frac{1}{32}< \frac{1}{30};\frac{1}{33}< \frac{1}{30};...;\frac{1}{40}< \frac{1}{30}\)

\(\Rightarrow A=\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}< \frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\)

\(\Rightarrow A< \frac{10}{30}=\frac{1}{3}(4)\)

Từ 3 và 4 \(\Rightarrow A< \frac{1}{2}+\frac{1}{3}=\frac{3}{6}+\frac{2}{6}=\frac{5}{6}\)

\(\Rightarrow A< \frac{5}{6}\)

Như vậy : \(\frac{7}{12}< A< \frac{5}{6}(đpcm)\)

Ta có:

\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}< \frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}\)

Mà \(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{1}{4}.4=1\)

=>\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}< 1\) (1)

\(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}< \frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\)Mà \(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{1}{8}.8=1\) 

=> \(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}< 1\)   (2)

Từ (1) và (2)

=> A=\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+...+\frac{1}{14}+\frac{1}{15}< 1+1\)

=> A<2

ê bài này ở đâu tek

dễ mà ngân

Ta có:

1/2 + 1/3 + 1/4 + ... + 1/15 + 1/16 = (1/2 + 1/3 + 1/4 + 1/5) + (1/6 + 1/7 + 1/8) + (1/9 + 1/10 + 1/11) + (1/12 + 1/13 + 1/14) + (1/15 + 1/16)

Vì 1/6 + 1/7 + 1/8 < 3x 1/6 = 1/2

   1/9 + 1/10 + 1/11 <3x1/9 = 1/3

   1/12 + 1/13 +1/14 < 3x1/12 = 1/4

   1/15 + 1/16 < 3 x 1/15 = 1/5

Nên A < 2 x (1/2 + 1/3 + 1/4 + 1/5) < 2 x (1/2 + 1/2 + 1/4 + 1/4) =3 (1)

Lập luận tương tự có:

A = ( 1/2 + 1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + 1/10 + 1/11 + 1/12) + (1/13 + 1/14 + 1/15 + 1/16) > (1/2 + 1/3 + 1/4) + 4 x 1/8 + 4 x 1/ 12 + 4 x 1/16

Hay A > 2 x (1/2 + 1/3 + 1/4) > 2 x (1/2 + 1/4 + 1/4) = 2 (2)

Từ (1) và (2) ta có 2 < A < 3. Vậy A không phải là số tự nhiên.