Ta có :

\(\dfrac{1}{199}+\dfrac{2}{198}+...+\dfrac{198}{2}+\dfrac{199}{1}\)

\(=\left(\dfrac{1}{199}+1\right)+\left(\dfrac{2}{198}+1\right)+...+\left(\dfrac{198}{2}+1\right)\left(\dfrac{199}{1}+1\right)-199\)\(=\dfrac{200}{199}+\dfrac{200}{199}+...+\dfrac{200}{2}+200-199\)

\(=\dfrac{200}{199}+\dfrac{200}{198}+...+\dfrac{200}{2}+\dfrac{200}{200}\)

\(=200\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{200}\right)\)

\(=200.A\)

\(\Rightarrow\dfrac{A}{B}=\dfrac{1}{200}\)

Ta có:\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}>\dfrac{1}{200}+\dfrac{1}{200}+...+\dfrac{1}{200}=\dfrac{100}{200}=\dfrac{1}{2}\)

Lại có:

\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}< \dfrac{1}{101}+\dfrac{1}{101}+...+\dfrac{1}{101}=\dfrac{100}{101}\)

Vậy ...

Những dãy trên đều có 100 số hạng.

\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{199}-\dfrac{1}{200}\)

\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{199}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+..+\dfrac{1}{200}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{199}+\dfrac{1}{200}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{200}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{199}+\dfrac{1}{200}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)\)

\(=\dfrac{1}{101}+...+\dfrac{1}{199}+\dfrac{1}{200}\)

Ta có: \(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{199\cdot200}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{199}-\dfrac{1}{200}\)

\(=\dfrac{1}{2}-\dfrac{1}{200}\)

\(=\dfrac{100-1}{200}=\dfrac{99}{200}< \dfrac{100}{200}=\dfrac{1}{2}\)(đpcm)

Ta có: \(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{199\cdot200}\)

          \(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\cdot\cdot\cdot+\dfrac{1}{199}-\dfrac{1}{200}\)

          \(=\dfrac{1}{2}-\dfrac{1}{200}< \dfrac{1}{2}\)  (Đpcm)

Ta có:\(C=\dfrac{1}{2}.\dfrac{3}{4}.....\dfrac{199}{200}\)

\(\Rightarrow C< \dfrac{2}{3}.\dfrac{4}{5}.....\dfrac{200}{201}\)

\(\Rightarrow C^2< \dfrac{2}{3}.\dfrac{4}{5}.....\dfrac{200}{201}.\dfrac{1}{2}.\dfrac{3}{4}.....\dfrac{199}{200}\)

\(\Rightarrow C^2< \dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.....\dfrac{199}{200}.\dfrac{200}{201}\)

\(\Rightarrow C^2< \dfrac{1}{201}\) (đpcm)

Ta có :

\(C=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{199}{200}< \dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{200}{201}\)

\(\Rightarrow C^2< \dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{199}{200}.\dfrac{200}{201}\)

\(\Rightarrow C^2< \dfrac{1.2.3.4....199.200}{2.3.4.5....200.201}=\dfrac{1}{201}\)

\(\Rightarrow\left(đpcm\right)\)

Câu a :

Áp dụng BĐT \(\dfrac{1}{\sqrt{ab}}>\dfrac{2}{a+b}\left(a\ne b;a,b>0\right)\) ta có :

\(\dfrac{1}{\sqrt{1.1998}}>\dfrac{2}{1+1998}=\dfrac{2}{1999}\)

\(\dfrac{1}{\sqrt{2.1997}}>\dfrac{2}{2+1997}=\dfrac{2}{19999}\)

.......................................................

\(\dfrac{1}{\sqrt{1998.1}}>\dfrac{2}{1998+1}=\dfrac{2}{1999}\)

Cộng tất cả vế với nhau ta được : \(P>2.\dfrac{1998}{1999}\)

\(\Rightarrowđpcm\)

Câu a, b sao tính chất cái cuối khác những cái còn lại thế. Vậy sao biết tới đâu thì nó dừng.