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)

Ta có:

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

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

Bài toán phụ 1:

Ta có:

1/13<1/12

1/14<1/12

1/15<1/12

=>1/13+1/14+1/15<1/12x3=1/4 (1)

Bài toán phụ 2:

Ta có:

1/61<1/60

1/62<1/60

1/63<1/60

=>1/61+1/62+1/63<1/60x3=1/20 (2)

Từ (1) và (2), ta có:

1/5+1/13+1/14+1/15+1/61+1/62+1/63<1/5+1/4+1/20

1/5+1/13+1/14+1/15+1/61+1/62+1/63<4/20+5/20+1/20

1/5+1/13+1/14+1/15+1/61+1/62+1/63<9/20<1/2

=>1/5+1/13+1/14+1/15+1/61+1/62+1/63<1/2

Ta có :

\(\frac{1}{13}< \frac{1}{12};\frac{1}{14}< \frac{1}{12};\frac{1}{15}< \frac{1}{12}\Rightarrow\frac{1}{13}+\frac{1}{14}+\frac{1}{15}< \frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{1}{4}\)

\(\frac{1}{61}< \frac{1}{60};\frac{1}{62}< \frac{1}{60};\frac{1}{63}< \frac{1}{60}\Rightarrow\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{60}+\frac{1}{60}+\frac{1}{60}=\frac{1}{20}\)

\(\Rightarrow D=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)< \frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{1}{2}\)

Vậy \(D< \frac{1}{2}\)

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

Nhận xét: \(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}< \frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\)

\(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{60}+\frac{1}{60}+\frac{1}{60}=\frac{3}{60}=\frac{1}{20}\)

\(\Rightarrow D< \frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{1}{2}\)

Vậy D < 1/2

Ta có : 

\(\frac{1}{13}< \frac{1}{12}\)

\(\frac{1}{14}< \frac{1}{12}\)

\(\frac{1}{15}< \frac{1}{12}\)

\(\Rightarrow\frac{1}{13}+\frac{1}{14}+\frac{1}{15}< \frac{1}{12}+\frac{1}{12}+\frac{1}{12}=3\cdot\frac{1}{12}=\frac{1}{4}\) (1)

Ta cũng có :

\(\frac{1}{61}< \frac{1}{60}\)

\(\frac{1}{62}< \frac{1}{60}\)

\(\frac{1}{63}< \frac{1}{60}\)

\(\Rightarrow\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{60}+\frac{1}{60}+\frac{1}{60}=3\cdot\frac{1}{60}=\frac{1}{20}\) (2)

Từ (1) ; (2) \(\Rightarrow S=\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{1}{2}\)

=> S < \(\frac{1}{2}\) (đpcm)

sao tui ko thíck làm thui nhé

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

suy ra S<1/5+1/12.3+1/60.3

S<1/5+1/4+1/20

S<1/2

Dãy số này không viết theo qui luật nên bạn chỉ có cách là cộng tất cả các số hạng vào rồi xét nó bé hơn \(\frac{1}{2}\)

câu trả lời của bạn lê phương thảo đúng rồi

\(A=\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)

Bạn xem lại đề.

Lấy máy tính bấn tổng kia thì bé hơn 1/2. Xem lại đề