cho A = 1/31 + 1/32 + 1/33 + ... + 1/60 CMR : A > 7/12
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 1/31 + 1/32 + 1/33 + ... + 1/60
=> A = (1/31 + 1/32 + ... + 1/45) + (1/46 + 1/47 + ... 1/60) > (1/45) x 15 + (1/60) x 15
=> A > 1/3 + 1/4 = 7/12
Vậy A > 7/12 (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)
Mà : (1/31+1/32+1/33+...+1/40) < 1/31 x 10 = 10/30 = 1/3 (gồm 10 số hạng)
Tương tự : (1/41 + 1/42 + ...+ 1/50) < 1/4 ; (1/51 + 1/52+...+1/59+1/60) < 1/5
Mà A = (1/3 + 1/4 + 1/5) = 47/60 > 7/12
Vậy A >7/12
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 1/31 + 1/32 + ... + 1/60
A = (1/31 + 1/32 + ... + 1/40) + (1/41 + 1/42 + ... + 50) + (1/51 + 1/52 + ... + 1/60)
A > 1/40 × 10 + 1/50 × 10 + 1/60 × 10
A > 1/4 + 1/5 + 1/6
A > 1/4 + 1/6 + 1/6
A > 1/4 + 1/3
A > 7/12
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
A: có 30 số hạng không đủ
phải chia nhỏ ra
\(A=\left(\frac{1}{31}+...+\frac{1}{36}\right)+\left(\frac{1}{37}+..+\frac{1}{48}\right)+\left(\frac{1}{49}+..+\frac{1}{60}\right)\)
\(A>\left(\frac{6}{36}\right)+\left(\frac{12}{48}\right)+\left(\frac{12}{60}\right)=\frac{3}{12}+\frac{3}{12}+\frac{1}{12}=\frac{7}{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)
\(=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+...+\frac{1}{60}\right)\)
\(< \left(\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\right)+\left(\frac{1}{40}+...+\frac{1}{40}\right)+\left(\frac{1}{50}+...+\frac{1}{50}\right)\)
\(=\frac{10}{30}+\frac{10}{40}+\frac{10}{50}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}=\frac{47}{60}< \frac{48}{60}=\frac{4}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)
Mà : (1/31+1/32+1/33+...+1/40) > 1/40 x 10 = 1/4 (gồm 10 số hạng)
Tương tự : (1/41 + 1/42 + ...+ 1/50) > 1/5 ; (1/51 + 1/52+...+1/59+1/60) > 1/6
S > 1/4 + 1/5 + 1/6.
Trong khi đó (1/4 + 1/5 + 1/6) > 3/5
=>S > 3/5 (1)
S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)
Mà : (1/31+1/32+1/33+...+1/40) < 1/31 x 10 = 10/30 = 1/3 (gồm 10 số hạng)
=> S < 4/5 (2)
Từ (1) và (2) => 3/5 <S<4/5
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)
\(=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{45}\right)+\left(\frac{1}{46}+...+\frac{1}{60}\right)>\frac{1}{45}.15+\frac{1}{60}.15=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
=>đpcm
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