Ta có: \(S=\frac{1}{5^2}+\frac{2}{5^3}+\cdots+\frac{99}{5^{100}}\)

=>\(5S=\frac15+\frac{2}{5^2}+\cdots+\frac{99}{5^{99}}\)

=>\(5S-S=\frac15+\frac{2}{5^2}+\cdots+\frac{99}{5^{99}}-\frac{1}{5^2}-\frac{2}{5^3}-\cdots-\frac{99}{5^{100}}=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)

=>\(4S=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)

Ta có: \(A=\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}\)

=>\(5A=\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{98}}\)

=>\(5A-A=\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{98}}-\frac{1}{5^2}-\frac{1}{5^3}-\cdots-\frac{1}{5^{99}}\)

=>\(4A=\frac15-\frac{1}{5^{99}}=\frac{5^{98}-1}{5^{99}}\)

=>\(A=\frac{5^{98}-1}{4\cdot5^{99}}\)

Ta có: \(4S=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)

\(=\frac15+\frac{5^{98}-1}{4\cdot5^{99}}-\frac{99}{5^{100}}=\frac15+\frac{5^{99}-5-396}{4\cdot5^{100}}=\frac15+\frac{1}{4\cdot5}-\frac{401}{4\cdot5^{100}}\)

=>\(4S<\frac15+\frac{1}{20}=\frac{4}{20}+\frac{1}{20}=\frac{5}{20}=\frac14\)

hay S<1/16

Ta có: \(S=\frac{1}{5^2}+\frac{2}{5^3}+\cdots+\frac{99}{5^{100}}\)

=>\(5S=\frac15+\frac{2}{5^2}+\cdots+\frac{99}{5^{99}}\)

=>\(5S-S=\frac15+\frac{2}{5^2}+\cdots+\frac{99}{5^{99}}-\frac{1}{5^2}-\frac{2}{5^3}-\cdots-\frac{99}{5^{100}}=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)

=>\(4S=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)

Ta có: \(A=\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}\)

=>\(5A=\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{98}}\)

=>\(5A-A=\frac15+\frac{1}{5^2}+\cdots+\frac{1}{5^{98}}-\frac{1}{5^2}-\frac{1}{5^3}-\cdots-\frac{1}{5^{99}}\)

=>\(4A=\frac15-\frac{1}{5^{99}}=\frac{5^{98}-1}{5^{99}}\)

=>\(A=\frac{5^{98}-1}{4\cdot5^{99}}\)

Ta có: \(4S=\frac15+\frac{1}{5^2}+\frac{1}{5^3}+\cdots+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)

\(=\frac15+\frac{5^{98}-1}{4\cdot5^{99}}-\frac{99}{5^{100}}=\frac15+\frac{5^{99}-5-396}{4\cdot5^{100}}=\frac15+\frac{1}{4\cdot5}-\frac{401}{4\cdot5^{100}}\)

=>\(4S<\frac15+\frac{1}{20}=\frac{4}{20}+\frac{1}{20}=\frac{5}{20}=\frac14\)

hay S<1/16

* Bỏ ngoặc vuông đi :( 

\(\text{Ta có:}\)

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

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

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

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

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

\(\text{Ta có:}\)

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

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

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

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

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?

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

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)