Question

Chứng minh tổng sau đây không phải là số tự nhiên \(S=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{15}+\frac{1}{16}\)

Answer 1

1/16+1/2=9/16 không phải là số tự nhiên

 

Answer 2

Có thể chứng minh được S>2 đó!

Answer 3

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.

Answer 4

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.

Answer 5

\(S=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}\right)>\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\text{ }\text{[}4\left(\frac{1}{8}+\frac{1}{12}+\frac{1}{16}\right)]\)
  \(=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)\) 
  \(=2\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)>2\left(\frac{1}{2}+2.\frac{1}{4}\right)=2\)  \(\left(1\right)\)
\(S=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)+\left(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}\right)+\left(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}\right)+\left(\frac{1}{15}+\frac{1}{16}\right)< \left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)+\text{[}4\left(\frac{1}{8}+\frac{1}{12}+\frac{1}{16}\right)\text{]}\)
                        \(=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)\)
                        \(=2\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)< 2\left[2\left(\frac{1}{2}+\frac{1}{4}\right)\right]=3\)\(\left(2\right)\)
Từ \(\left(1\right)\), \(\left(2\right)\)ta có: \(2< S< 3\). Vậy \(S\inℕ\).