Đặt  \(A=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=30.\frac{1}{60}=\frac{1}{2}\)

       \(B=\frac{1}{61}+\frac{1}{62}+\frac{1}{63}+...+\frac{1}{90}>\frac{1}{90}+\frac{1}{90}+\frac{1}{90}+...+\frac{1}{90}=30.\frac{1}{90}=\frac{1}{3}\)

\(=>Q=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{90}=A+B>\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)

Vậy \(Q>\frac{5}{6}\)

Đã trả lời ở đâu đó rồi (chi tiết)

-Nhận xét, phân tích bài toán:

So sánh với (5/6) =>rút gọn vế trái thành một phân số có mẫu số bằng 6

=> ta chọn số hạng có mẫu số là bội số của 6 để gom lại.

\(\frac{1}{31}+..+\frac{1}{36}>\frac{1}{36}+..+\frac{1}{36}=\frac{6}{36}=\frac{1}{6}\)

\(\frac{1}{37}+...+\frac{1}{42}>\frac{1}{42}+..+\frac{1}{42}=\frac{6}{42}=\frac{1}{7}\)

..........

\(\frac{1}{83}+..+\frac{1}{90}=\frac{1}{90}+...+\frac{1}{90}=\frac{6}{90}=\frac{1}{15}\)

Như vậy sau bước 1 rút vê trái về còn \(\frac{1}{6}+\frac{1}{7}...+\frac{1}{15}\)

Rút gọn tiếp vẫn theo cách trên

\(\frac{1}{7}+..+\frac{1}{12}>\frac{1}{12}+..+\frac{1}{12}=\frac{6}{12}=\frac{3}{6}\)

\(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}>\frac{1}{18}+\frac{1}{18}+\frac{1}{18}=\frac{1}{6}\)

\(VT=\left(\frac{1}{31}+..+\frac{1}{90}\right)>\left(\frac{1}{6}+\frac{3}{6}+\frac{1}{6}\right)=\frac{5}{6}=VP\)

Hay thật!

Ta có: A=1/11+1/12+1/13+...+1/30

            =(1/11+1/12+1/13+..+1/20)+(1/21+1/22+1/23+...+1/30)

\(\Rightarrow\)A<(1/10+1/10+1/10+...+1/10)+(1/20+1/20+1/20+...1/20)

\(\Rightarrow\)A<(1/10)*10+(1/20)*10

\(\Rightarrow\)A<1+1/2

\(\Rightarrow\)A<3/2<11/6

cam on ban rat nhieu

????????????

Đặt S =\(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{1458}+\frac{1}{4374}\)

3S = \(3\times\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{1458}+\frac{1}{4374}\right)\)

3S \(=\frac{3}{2}+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{486}+\frac{1}{1458}\)

3S - S \(=\left(\frac{3}{2}+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{486}+\frac{1}{1458}\right)-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{1458}+\frac{1}{4374}\right)\)

2S = \(\frac{3}{2}+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{486}+\frac{1}{1458}-\frac{1}{2}-\frac{1}{6}-...-\frac{1}{1458}-\frac{1}{4374}\)

2S = \(\frac{3}{2}-\frac{1}{4374}\)

2S = \(\frac{3280}{2187}\)

\(\Rightarrow S=\frac{3280}{2187}:2=\frac{4373}{8748}\)

\(=1-\frac{1}{2}+\frac{1}{2}+\frac{1}{3}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1+\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{99}+\frac{1}{99}\right)-\frac{1}{100}\)

\(=1+0+0+...+0-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}< 11\)

Vậy : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}< 11\)

=1/2-1/3+1/3-1/4+...+1/99-1/100

=1/2-1/100

=50/100-1/100

=49/100<1

=> dãy trên < 1 đđcm