Ta có : \(\frac{1}{2}+\frac{2}{3}+..+\frac{99}{100}\)

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

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

= \(1\times100-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{100}\right)\)

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

=> 100-(1+1/2+1/3+...+1/100) = 1/2+2/3+3/4+...+99/100

Bạn cố giải cho mình dễ hiểu hơn ko?

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}< \frac{1}{2}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}< \frac{1}{2}\)

\(=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\left(đpcm\right)\)

\(2^2< 2.3\Rightarrow\dfrac{1}{2^2}>\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\)

Tương tự: \(\dfrac{1}{3^2}>\dfrac{1}{3}-\dfrac{1}{4}\) ; \(\dfrac{1}{4^2}>\dfrac{1}{4}-\dfrac{1}{5}\) ; ....; \(\dfrac{1}{100^2}>\dfrac{1}{100}-\dfrac{1}{101}\)

Do đó:

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{100}-\dfrac{1}{101}\)

\(\Leftrightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}>\dfrac{1}{2}-\dfrac{1}{101}\)

\(\Leftrightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}>\dfrac{99}{202}\)

= (1+1/3+1/5+…+1/99)-(1/2+1/4+….+1/100)

= (1+1/2+1/3+…+1/100)-2(1/2+1/4+1/6+…+1/100)

= (1+1/2+1/3+…+1/100)-(1+1/2+1/3+…+1/50)

=1/51+1/52+…+1/100=VP (đpcm)

= (1+1/3+1/5+…+1/99)-(1/2+1/4+….+1/100)

= (1+1/2+1/3+…+1/100)-2(1/2+1/4+1/6+…+1/100)

= (1+1/2+1/3+…+1/100)-(1+1/2+1/3+…+1/50)

=1/51+1/52+…+1/100=VP (đpcm)

* Bỏ ngoặc vuông đi :( 

\(\text{Ta có:}\)

\(200-\left(3+\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100}\right)\)

\(\rightarrow200-2-\left(1+\frac{2}{3}+...+\frac{2}{100}\right)\)

\(\rightarrow198-\left(1+\frac{2}{3}+...+\frac{2}{100}\right)\)

\(\rightarrow198-\left(1+\frac{2}{3}+...+\frac{2}{100}\right)\)

\(\rightarrow2.[99-\left(\frac{1}{2}-\frac{1}{3}+...+\frac{1}{100}\right)]\)     \(\left(1\right)\)

\(\text{Ta có:}\)

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

\(\text{Rút}\)\(\left(1\right)\)\(\text{ra có 99 số}\)

\(\rightarrow99-\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)     \(\left(2\right)\)

\(\text{Từ}\)\(\left(1\right)\)\(\text{và}\)\(\left(2\right)\)\(\Rightarrow\)\(200-\left(3+\frac{2}{3}+\frac{2}{4}+\frac{2}{5}+...+\frac{2}{100}\right):\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\right)=2\)

VÌ \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2};\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3};...........;\frac{1}{99^2}=\frac{1}{99\cdot99}< \frac{1}{99\cdot100}\)

\(\Rightarrow S< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+.....+\frac{1}{99\cdot100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)\(=1-\frac{1}{100}< 1\)\(\Rightarrow S< 1\)

VÌ \(\frac{1}{2\cdot3}< \frac{1}{2\cdot2};.....;\frac{1}{98\cdot99}< \frac{1}{99\cdot99}\)

\(\Rightarrow\)\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+......+\frac{1}{98\cdot99}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}=\frac{50}{100}-\frac{1}{100}=\frac{49}{100}< S\)

\(\Rightarrow\frac{49}{100}< S< 1\)

\(K\)\(mk\)\(nha\)