a,\(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)

\(=>5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)

\(=>5A-A=1-\frac{1}{5^{100}}=>A=\frac{1-\frac{1}{5^{100}}}{4}\)

b, Ta có \(1-\frac{1}{5^{100}}< 1=>\frac{1-\frac{1}{5^{100}}}{4}< \frac{1}{4}\)hay \(A< \frac{1}{4}\)

Gọi phân số 10^2014+1/10^2015+1 là A

Gọi phân số 10^2015+1/10^2016+1

Xét thấy B = 10^2015+1/10^2016+1 là phân số nhỏ hơn 1

=> theo tính chất : Nếu a/b<1 thì a/b<(a+n)/(b+n) (a,b,n thuộc N ;b;n khác 0)

=> B = (10^2015+1)/(10^2016+1) < (10^2015+1+9)/(10^2016+1+9) = (10^2015+10/10^2016+10)

=> B < 10.(10^2014+1)/10.(10^2015+1)

=> B < 10^2014+1/10^2015+1 = A (cùng bớt 10 ở tử và mẫu)

 Vậy B < A                                   

A- 1 = \(\frac{10^{2015}-1-\left(10^{2016}-1\right)}{10^{2016}-1}=\frac{-9.10^{2015}}{10^{2016}-1}=\frac{-90.10^{2014}}{10^{2016}-1};\)

B- 1 = \(\frac{10^{2014}+1-\left(10^{2015}+1\right)}{10^{2015}+1}=\frac{-9.10^{2014}}{10^{2015}+1};\)

xét \(\frac{A-1}{B-1}=\frac{-90.10^{2014}}{10^{2016}-1}:\frac{-9.10^{2014}}{10^{2015}+1}=\frac{10\left(10^{2015}+1\right)}{10^{2016}-1}=\frac{10^{2016}+10}{10^{2016}-1}>1\)

=> A-1 > B-1 => A > B

\(A=\frac{10^{2015}-1}{10^{2016}^{ }-1}=\frac{10^{2015}}{10^{2016}}=\frac{1}{1},B=\frac{10^{2014}-1}{10^{2015}-1}=\frac{10^{2014}}{10^{2015}}=\frac{1}{1}A=B\Rightarrow\)

Ta có: \(10A=10.\left(\frac{10^{2014}+1}{10^{2015}+1}\right)=\frac{10^{2015}+10}{10^{2015}+1}=\frac{10^{2015}+1+9}{10^{2015}+1}=1+\frac{9}{10^{2015}+1}\)

\(10B=10.\left(\frac{10^{2015}+1}{10^{2016}+1}\right)=\frac{10^{2016}+10}{10^{2016}+1}=\frac{10^{2016}+1+9}{10^{2016}+1}=1+\frac{9}{10^{2016}+1}\)

Vì 1 = 1; 9 = 9 ta so sánh mẫu:

Ta có: 10²⁰¹⁵ < 10²⁰¹⁶ => 10²⁰¹⁵+1 < 10²⁰¹⁶+1

=> \(1+\frac{9}{10^{2015}+1}>1+\frac{9}{10^{2016}+1}\)

=> 10A > 10B

=> A > B.