Ta có 

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

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

\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

=> S < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

S < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(S< 1-\dfrac{1}{100}< 1\)(do 1/100 >0)

ĐPcm

Giải:

\(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}\) 

Ta có:

\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\) 

\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\) 

\(\dfrac{1}{4^2}=\dfrac{1}{4.4}< \dfrac{1}{3.4}\) 

\(...\) 

\(\dfrac{1}{99^2}=\dfrac{1}{99.99}< \dfrac{1}{98.99}\) 

\(\dfrac{1}{100^2}=\dfrac{1}{100.100}< \dfrac{1}{99.100}\) 

\(\Rightarrow S< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\) 

\(\Rightarrow S< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\) 

\(\Rightarrow S< \dfrac{1}{1}-\dfrac{1}{100}< 1\) 

\(\Rightarrow S< 1\) 

Vậy S < 1.

ta có: \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\)

\(\Rightarrow\frac{1}{3}A=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{100}}+\frac{1}{3^{101}}\)

\(\Rightarrow A-\frac{1}{3}A=\frac{1}{3}-\frac{1}{3^{101}}< \frac{1}{3}\)

\(\Rightarrow\frac{2}{3}A< \frac{1}{3}\)

\(\Rightarrow A< \frac{1}{3}:\frac{2}{3}\)

\(\Rightarrow A< \frac{1}{2}\)

Ta có 

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

\(2A=1+\frac{2}{2}+\frac{3}{2^2}+...+\frac{99}{2^{98}}+\frac{100}{2^{99}}\)

Suy ra \(A=2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\frac{100}{2^{100}}\)

Đặt \(n=\frac{1}{2}\) thì \(A=1+n+n^2+...+n^{99}-\frac{100}{2^{100}}\)

Xét \(B=1+n+n^2+...+n^{99}\Leftrightarrow B.n=n+n^2+n^3+...+n^{100}\)

\(\Leftrightarrow B.n=\left(1+n+n^2+...+n^{99}\right)+\left(n^{100}-1\right)\)

\(\Leftrightarrow B.n=B+n^{100}-1\Leftrightarrow B\left(n-1\right)=n^{100}-1\Leftrightarrow B=\frac{n^{100}-1}{n-1}\)

Suy ra \(A=\frac{\frac{1}{2^{100}}-1}{\frac{1}{2}-1}-\frac{100}{2^{100}}=2\left(1-\frac{1}{2^{100}}\right)-\frac{100}{2^{100}}=-\frac{102}{2^{100}}+2< 2\)

Vậy A < 2

#)Giải :

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

\(A=\frac{2}{9^1}+\frac{2}{9^2}+\frac{2}{9^3}+...+\frac{2}{9^{50}}\)

\(\Rightarrow2A=1+\frac{2}{9}+\frac{2}{9^1}+\frac{2}{9^2}+\frac{2}{9^3}+...+\frac{2}{9^{49}}\)

\(\Rightarrow2A-A=A=\left(1+\frac{2}{9}+\frac{2}{9^1}+\frac{2}{9^2}+\frac{2}{9^3}+...+\frac{2}{9^{49}}\right)-\left(\frac{2}{9^1}+\frac{2}{9^2}+\frac{2}{9^3}+...+\frac{2}{9^{50}}\right)\)

\(\Rightarrow A=1+\frac{2}{9}-\frac{2}{9^{50}}=\frac{11}{9}-\frac{2}{9^{50}}\)

Có lẽ đúng .........................

#)So sánh thì tự làm nhé !

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

\(\frac{1}{3}A=\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-\frac{1}{3^5}+...+\frac{1}{3^{100}}-\frac{1}{3^{101}}\)

\(\Rightarrow A+\frac{1}{3}A=\frac{1}{3^1}+\left(\frac{-1}{3^{101}}\right)=\frac{1}{3^1}-\frac{1}{3^{101}}\)

\(\Rightarrow A\left(1+\frac{1}{3}\right)=\frac{1}{3^1}-\frac{1}{3^{101}}\)

\(\Rightarrow\frac{4}{3}A=\frac{1}{3^1}-\frac{1}{3^{101}}\)

          \(A=\left(\frac{1}{3^1}-\frac{1}{3^{101}}\right):\frac{4}{3}\)

          \(A=\left(\frac{1}{3^1}-\frac{1}{3^{101}}\right).\frac{3}{4}\)

          \(A=\frac{1}{3^1}.\frac{3}{4}-\frac{1}{3^{101}}.\frac{3}{4}\)

         \(A=\frac{1}{4}-\frac{1}{3^{100}.4}< \frac{1}{4}< \frac{1}{3}\)

\(\Rightarrow A< \frac{1}{3}\)

      Vậy \(A< \frac{1}{3}\)

Ta có: \(\frac{1}{2}A=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{100}{2^{101}}\)

\(A-\frac{1}{2}A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}-\frac{100}{2^{101}}\)

Ta có: \(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}=1-\frac{1}{2^{100}}< 1\)

\(\Rightarrow\frac{1}{2}A< 1-\frac{100}{2^{101}}\)

\(\Rightarrow A< 2-\frac{200}{2^{101}}< 2\)

Vậy A<2

2 vế bằng nhau

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

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

100 = 1 + 1/2 + 1/2 + 1/3 + 2/3 + ... + 1/100 + 99/100 (cùng cộng 2 vế với (- 1-1/2-1/3-...-1/100)

100 = 1 + 1 + 1 + ... + 1 (100 số hạng)

100 = 100

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

Cảm ơn bạn!

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

\(A< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

\(A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

\(A< 1+1-\frac{1}{100}\)

\(A< 2-\frac{1}{100}< 2\)

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

\(A< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

\(A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

\(A< 1+1-\frac{1}{100}\)

\(A< 2-\frac{1}{100}< 2\)