Ta có : 

S = \(\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)

a) Ta có:

S = 1/5 + 1/13 + 1/14 + 1/15 + 1/61 + 1/62 + 1/63

Ta thấy:

1/13 < 1/12 ; 1/14 < 1/12 ; 1/15 < 1/12

=> 1/13 + 1/14 + 1/15 < 1/12 + 1/12 + 1/12 = 1/12 . 3 = 1/4  (1)

1/61 < 1/60 ; 1/62 < 1/60 ; 1/63 < 1/60

=> 1/61 + 1/62 + 1/63 < 1/60 + 1/60 + 1/60 = 1/60. 3 = 1/20  (2)

 Từ (1) và (2)

=> 1/13 + 1/14 + 1/15 + 1/61 + 1/62 + 1/63 < 1/4 + 1/20

=>S =  1/5 + 1/13 + 1/14 + 1/15 + 1/61 + 1/62 + 1/63 < 1/4 + 1/20 + 1/5 = 5/20 + 1/20 + 4/20 = 10/20 = 1/2 (ĐPCM)

b) Ta có:

\(P=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}\)

\(2P=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\)

\(2P-P=1+\frac{1}{2}-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^2}+...+\frac{1}{2^{19}}-\frac{1}{2^{19}}-\frac{1}{2^{20}}\)

\(P=1-\frac{1}{2^{20}}< 1\)

=> P < 1

Ta có: 

\(\frac{1}{5}=\frac{1}{5}\)

\(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}

Ta có: \(S=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)

\(\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}

Ta có:

S=1/5+(1/13+1/14+1/15)+(1/61+1/62+1/63)<1/5+1/12.3+1/60.3

=>S<1/5+1/4+1/20=10/20

Hay S<1/2

Ta có: \(A=\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\)

\(A=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{62}+\frac{1}{62}+\frac{1}{63}\right)\)

\(A=\frac{1}{5}+\frac{1}{15}.3+\frac{1}{63}.3\)

\(A=\frac{1}{5}+\frac{1}{5}+\frac{1}{21}\)

\(A=\frac{47}{105}\)

Mà: \(\frac{47}{105}< \frac{47}{94}=\frac{1}{2}\)

Nên \(A=\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{2}\)