\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) ta có : 

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

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

\(A< 1-\frac{1}{100}=\frac{99}{100}< 1\)

Vậy \(A< 1\)

Chúc bạn học tốt ~ 

\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}\) ta có : 

\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)

\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)

\(A< 1-\frac{1}{2010}=\frac{2009}{2010}< 1\)

\(\Rightarrow\)\(A< 1\) ( đpcm ) 

Vậy \(A< 1\)

Chúc bạn học tốt ~ 

2A=1+1/2+1/2^2+1/2^3+...+1/2^99

-A=    1/2+1/2^2+1/2^3+...+1/2^99+1/2^100

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A=1-1/2^100

A=2^100-1/2^100<1(dpcm)

B), B=2/1.2 +22.3 +23.4 +...+299.100 <2 =

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

=1-1/100

=99/100 

vì 99/100<2 nên B=2/1.2+2/2.3+2/3.4+......+2/99.100<2

a,1/10²+1/11²+1/12²+...+1/100²<1/9.10+1/10.11+1/11.12+...+1/99.100=1/9-1/10+1/10-1/11+...+1/99-1/100

                                                                                                    =1/9-1/100=91/900<3/4

Vậy 1/10²+1/11²+1/12²+...+1/100²<3/4

b,1/2²+1/3²+1/4²+...+1/100²<1/1.2+1/2.3+1/3.4+...+1/99.100=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100

                                                                                        =1-1/100=99/100

Vậy 1/2²+1/3²+1/4²+...+1/100²<99/100

c,1/2²+1/3²+1/4²+...+1/100²<1/2²+(1/2.3+1/3.3+...+1/99.100)=1/4+(1/2-1/3+1/3-1/4+...+1/99-1/100)

                                                                                       =1/4+(1/2-1/100)=1/4+49/100=74/100<3/4=75/100

Vậy 1/2²+1/3²+1/4²+...+1/100²<3/4

a) \(A=\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2019}}\)

\(5A=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2018}}\)

\(4A=5A-A=\frac{1}{5}-\frac{1}{5^{2019}}\)

\(A=\frac{1}{20}-\frac{1}{4.5^{2019}}< \frac{1}{20}< \frac{1}{2}\)

b)  Đề có sai không mà đằng cuối lại là \(\frac{1}{4^2}\)lặp lại lần nữa.
c) \(C=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)

\(2C=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}\)

\(3C=2C+C=1-\frac{1}{64}< 1\)

\(C< \frac{1}{3}\)

d) Xem lại đề nữa đi e, nếu trừ hai vế cho \(\frac{1}{3}\)thì vế trái > 0 > vế phải rồi
e)  \(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}>\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\)(10 số hạng)
                                                    \(=\frac{10}{50}=\frac{1}{5}\)

Tương tự: \(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}>\frac{1}{6}\)

\(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{70}>\frac{1}{7}\)

\(\frac{1}{71}+\frac{1}{72}+...+\frac{1}{80}>\frac{1}{8}\)

\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}>\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}=\frac{533}{840}>\frac{490}{840}=\frac{7}{12}\)

\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2006}}\)

\(\Rightarrow2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2005}}\)

\(2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^{2005}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2006}}\right)\)

\(2A=2-\frac{1}{2^{2006}}\)

\(\Rightarrow A=\frac{2-\frac{1}{2^{2006}}}{2}=1-\frac{1}{2^{2007}}\)

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

\(\Rightarrow3B=-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)

\(3B+B=\left(-\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}-\frac{1}{3^{101}}\right)+\left(-1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\right)\)

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

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

\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2005}}\)

\(2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2005}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2006}}\right)\)

\(A=2-\frac{1}{2^{2006}}\)

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

\(3B+B=\left(-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\right)+\left(-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{100}}-\frac{1}{3^{101}}\right)\)

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

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