Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{2019^2}\)

\(\Rightarrow A=\frac{1}{2^2}+\left(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2019^2}\right)\)

\(\Rightarrow A< \frac{1}{4}+\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2018.2019}\right)\)

\(\Rightarrow A< \frac{1}{4}+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+..+\frac{1}{2018}-\frac{1}{2019}\right)\)

\(\Rightarrow A< \frac{1}{4}+\left(\frac{1}{2}-\frac{1}{2019}\right)\)

\(\Rightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{2019}=\frac{3}{4}-\frac{1}{2019}< \frac{3}{4}\)

\(\Rightarrow A< \frac{3}{4}\)

đặt A=1/2^2+....+1/2019^2

vì 1/2^2+....+1/2019^2<1/1.2+1/2.3+....+1/2018.2019

=> A<1/1-1/2+1/2-1/3+.....+1/2018-1/2019

=> A<1-1/2019=2018/2019<3/4.

=> A<3/4. 

vậy 1/2^2+....+1/2019^2<3/4

Ta có: \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\)

\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}\)\(+...+\frac{1}{2018}-\frac{1}{2019}\)

\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< 1-\frac{1}{2019}\)

\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< 1-\frac{1}{2019}\)

\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2019^2}< \frac{2018}{2019}\)

Mà: \(\frac{3}{4}=\frac{2016}{2688}< \frac{2017}{2688}< \frac{2017}{2019}< \frac{2018}{2019}\)

\(\Rightarrow\frac{3}{4}< \frac{2018}{2019}\)

\(D=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2018}{4^{2018}}+\frac{2019}{4^{2019}}\)

\(\Rightarrow4D=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2018}{4^{2017}}+\frac{2019}{4^{2018}}\)

\(\Rightarrow4D-D=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2018}{4^{2017}}+\frac{2019}{4^{2018}}\)

\(-\frac{1}{4}-\frac{2}{4^2}-\frac{3}{4^3}-\frac{4}{4^4}-...-\frac{2018}{4^{2018}}-\frac{2019}{4^{2019}}\)

\(\Rightarrow3D=1+\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2018}}\right)-\frac{2019}{4^{2019}}\)

Đặt \(M=\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\frac{1}{4^4}+...+\frac{1}{4^{2018}}\)

\(\Rightarrow4M=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2017}}\)

\(\Rightarrow4M-M=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2017}}\)

\(-\frac{1}{4}-\frac{1}{4^2}-\frac{1}{4^3}-\frac{1}{4^4}-...-\frac{1}{4^{2018}}\)

\(\Rightarrow3M=1-\frac{1}{4^{2018}}\)

\(\Rightarrow M=\frac{1}{3}-\frac{1}{3.4^{2018}}\)

\(\Rightarrow3D=1+\frac{1}{3}-\frac{1}{3.4^{2018}}-\frac{2019}{4^{2019}}\)

\(\Rightarrow3D=\frac{4}{3}-\frac{1}{3.4^{2018}}-\frac{2019}{4^{2019}}< \frac{4}{3}\)

\(\Rightarrow D< \frac{4}{9}=\frac{40}{90}< \frac{45}{90}=\frac{1}{2}\left(đpcm\right)\)

a/

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

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

b/

\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2018}}\)

\(2B=3B-B=1-\frac{1}{3^{2019}}\Rightarrow B=\frac{1}{2}-\frac{1}{2.3^{2019}}< \frac{1}{2}\)

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2011^2}\)

\(\text{Vì}\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{2011^2}< \frac{1}{2010.2011}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2011^2}< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2010.2011}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2011^2}< \frac{1}{2^2}+\frac{1}{2}-\frac{1}{2011}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2011^2}< \frac{1}{4}+\frac{1}{2}-\frac{1}{2011}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2011^2}< \frac{3}{4}-\frac{1}{2011}< \frac{3}{4}\)

\(\Rightarrowđpcm\)

Phần C đề thiếu

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

\(\Rightarrow3D=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)

\(\Rightarrow3D-D=(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}})-\)\((\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}})\)

\(\Rightarrow2D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(\Rightarrow6D=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)

\(\Rightarrow6D-2D=3-\frac{101}{3^{99}}+\frac{100}{3^{100}}\)

\(\Rightarrow4D=3-\frac{203}{3^{100}}\)

\(\Rightarrow D=\frac{3}{4}-\frac{\frac{203}{3^{100}}}{4}< \frac{3}{4}\left(đpcm\right)\)

sửa rồi nhá bn

\(C=\frac{1}{2}-\frac{1}{2^2}+...+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)

\(\Rightarrow2C=1-\frac{1}{2}+...+\frac{1}{2^{98}}-\frac{1}{2^{99}}\)

\(\Rightarrow2C+C=(1-\frac{1}{2}+...+\frac{1}{2^{98}}-\frac{1}{2^{99}})+\)\((\frac{1}{2}-\frac{1}{2^2}+...+\frac{1}{2^{99}}-\frac{1}{2^{100}})\)

\(\Rightarrow3C=1-\frac{1}{100}\)

\(\Rightarrow C=\frac{1}{3}-\frac{1}{300}< \frac{1}{3}\left(đpcm\right)\)

Ta có: \(S=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2019!}=1+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2019!}\)

Đặt \(M=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{2019!}\)

\(\Rightarrow M< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}\)

\(\Rightarrow M< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)

\(\Rightarrow M< 1-\frac{1}{2019}=\frac{2019}{2019}-\frac{1}{2019}=\frac{2018}{2019}\)

\(\Rightarrow S< 1+\frac{2018}{2019}=\frac{2019}{2019}+\frac{2018}{2019}=\frac{4037}{2019}< 2\)

\(\Rightarrow S< 2\) ( ĐPCM )

Ta có : \(B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}\)

Mà \(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};...;\frac{1}{8^2}<\frac{1}{7.8}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}=1-\frac{1}{8}<1\)

Vậy B < 1

Ta có:\(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};......;\frac{1}{8^2}<\frac{1}{7.8}\) 

<=> B<\(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{7.8}\) 

<=> B<\(\frac{1}{1}-\frac{1}{2}+.......+\frac{1}{7}-\frac{1}{8}\) 

<=> B<\(1-\frac{1}{8}\) 

<=> B<\(\frac{7}{8}\) <1