We are going to have to get some stuff 

S = \(\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)+\left(\frac{1}{111}+...+\frac{1}{120}\right)+\left(\frac{1}{121}+...+\frac{1}{130}\right)\)

>  \(\frac{1}{110}.10+\frac{1}{120}.10+\frac{1}{130}.10=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}>\frac{1}{12}+\frac{2}{12}=\frac{1}{4}\) (Dễ có: \(\frac{1}{11}+\frac{1}{13}>\frac{2}{12}\))

=> S > \(\frac{1}{4}\) (1)

+) S = \(\left(\frac{1}{101}+\frac{1}{130}\right)+\left(\frac{1}{102}+\frac{1}{129}\right)+...+\left(\frac{1}{115}+\frac{1}{116}\right)\) (Có 15 cặp)

= \(\frac{231}{101.130}+\frac{231}{102.129}+...+\frac{231}{115.116}=231.\left(\frac{1}{101.130}+\frac{1}{102.129}+...+\frac{1}{115.116}\right)\)

ta có nhận xét: tích 101.130 có giá trị nhỏ nhất. thật vậy:

Xét 102.129 = (101 + 1).(130 - 1) = 101.130 - 101 + 130 -1 = 101.130 + 28 > 101.130

Tương tự, các cặp còn lại . Do đó, ta có \(\frac{1}{101.130}+\frac{1}{102.129}+...+\frac{1}{115.116}

Sao bạn học giỏi thế? 

BẠN HỌC GIOI THẾ

Refer

Chứng minh \(S< \dfrac{91}{330}\)

\(S=\left(\dfrac{1}{101}+\dfrac{1}{102}+.....+\dfrac{1}{110}\right)+\left(\dfrac{1}{111}+....+\dfrac{1}{120}\right)+\left(\dfrac{1}{121}+......+\dfrac{1}{130}\right)\)

\(S< \left(\dfrac{1}{100}+\dfrac{1}{100}......+\dfrac{1}{100}\right)+\left(\dfrac{1}{110}+....+\dfrac{1}{110}\right)+\left(\dfrac{1}{120}+....+\dfrac{1}{120}\right)\)

\(S< \dfrac{66+60+65}{660}\)

\(S< \dfrac{181}{660}< \dfrac{182}{660}\)

+ Hay \(S< \dfrac{91}{330}\left(1\right)\)

Chứng minh \(\dfrac{1}{4}< S\)

\(S>\left(\dfrac{1}{110}\right)+.....+\left(\dfrac{1}{110}\right)+\left(\dfrac{1}{120}\right)+.....+\left(\dfrac{1}{120}\right)+\left(\dfrac{1}{130}\right)+......+\left(\dfrac{1}{130}\right)\)

\(S>\dfrac{1}{110}.10+\dfrac{1}{120}.10+\dfrac{1}{130}.10=\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}\)

\(S>\dfrac{156+143+132}{1716}\)

+ Hay \(S>\dfrac{1}{4}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\dfrac{1}{4}< S< \dfrac{91}{330}\)

tk

Bạn tham khảo ở đây ạ =))

YaMate