Bạn thiếu đề rồi phải là trừ hay cộng j j chứ.

Xét:

`A+B=2+1/2+1/3+1/4+......+1/4026+1/3+1/5+1/7+......+1/4025`

`1/2+1/3+1/4+......+1/4026+1/3+1/5+1/7+......+1/4025>0`

`=>A+B>2`

Mà `1 2013/2014<2`

`=>A+B>1 2013/2014`

Lời giải:
$M=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{2014}{5^{2014}}$

$5M=1+\frac{2}{5}+\frac{3}{5^2}+...+\frac{2014}{5^{2013}}$

$\Rightarrow 4M=5M-M=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2013}}-\frac{2014}{5^{2014}}$
$4M+\frac{2014}{5^{2014}}=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2013}}$

$5(4M+\frac{2014}{5^{2014}})=5+1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2012}}$

$\Rightarrow 4(4M+\frac{2014}{5^{2014}})=5-\frac{1}{5^{2013}}$

$M=\frac{5}{16}-\frac{1}{16.5^{2013}-\frac{2014}{4.5^{2014}}$

1/3²< 1/2.3

1/4²< 1/3.4

...

1/100²< 1/99.100

=> 1/2² + 1/3² + 1/4² + ... + 1/100²< 1/2² + 1/2.3 + 1/3.4 + ... + 1/99.100

A < 1/4 + 1/2 -1/3 + 1/3 - 1/4 +... + 1/99 - 1/100

A < 1/4 + 1/2 -1/100 < 1/4 + 1/2 = 3/4

=> A < 3/4

Ta có `3A=1+1/3+....+1/3^99`

`=>3A-A=1-1/3^100`

`=>2A=1-1/3^100`

`=>A=1/2-1/(2.3^100)<1/2`

Hay `A<B`

\(A=\dfrac{1}{3^1}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2023}}\)

\(A=\dfrac{1}{3}.\left(1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\right)\)

\(\Rightarrow3A=3.\dfrac{1}{3}.\left(1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\right)\)

\(\Rightarrow3A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\)

\(\Rightarrow3A-A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...\dfrac{1}{3^{2022}}-\left(\dfrac{1}{3^1}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2023}}\right)\)

\(\Rightarrow2A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...\dfrac{1}{3^{2022}}-\dfrac{1}{3^1}-\dfrac{1}{3^2}-\dfrac{1}{3^3}-...\dfrac{1}{3^{2022}}-\dfrac{1}{3^{2023}}\)

\(\Rightarrow2A=1-\dfrac{1}{3^{2023}}\)

\(\Rightarrow A=\dfrac{1}{2}\left(1-\dfrac{1}{3^{2023}}\right)\)

\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2}.\dfrac{1}{3^{2023}}< \dfrac{1}{2}\)

\(B=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{12}=\dfrac{4+3+1}{12}=\dfrac{8}{12}=\dfrac{2}{3}\)

mà \(\dfrac{2}{3}>\dfrac{1}{2}\) \(\left(\dfrac{2}{3}=\dfrac{4}{6}>\dfrac{1}{2}=\dfrac{3}{6}\right)\)

\(\Rightarrow A< B\)

       A =      \(\dfrac{1}{3}\)+ \(\dfrac{1}{3^2}\)+ \(\dfrac{1}{3^3}\)+............+\(\dfrac{1}{3^{2023}}\)

     3A = 1+ \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\)+...+\(\dfrac{1}{3^{2022}}\)

3A - A =  1 - \(\dfrac{1}{3^{2023}}\)

   2A   = 1 - \(\dfrac{1}{3^{2023}}\) < 1

      B =  \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\)+ \(\dfrac{1}{12}\)

      B  = \(\dfrac{4}{12}\) + \(\dfrac{3}{12}\) + \(\dfrac{1}{12}\)

     B   = \(\dfrac{8}{12}\)

     B   = \(\dfrac{2}{3}\) ⇒ 2B = \(\dfrac{4}{3}\) > 1 

2A < 2B ⇒ A < B 

\(A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2014}}\)

\(3A=3\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2014}}\right)\)

\(3A=3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{2013}}\)

\(3A-A=\left(3+1+...+\dfrac{1}{3^{2013}}\right)-\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^{2014}}\right)\)

\(2A=3-\dfrac{1}{3^{2014}}\Rightarrow A=\dfrac{3}{2}-\dfrac{\dfrac{1}{3^{2014}}}{2}< \dfrac{3}{2}\)

Vậy \(A< \dfrac{3}{2}\)

\(A=-\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{2014^2}\right)\)

\(A=\dfrac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2012\cdot2014\right)\left(2013\cdot2015\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2013\cdot2013\right)\left(2014\cdot2014\right)}\)

\(A=\dfrac{\left(1\cdot2\cdot3\cdot...\cdot2012\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2014\cdot2015\right)}{\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)}\)

\(A=\dfrac{1\cdot2015}{2014\cdot2}=\dfrac{2015}{4028}\)

Vì \(\dfrac{2015}{4028}>-\dfrac{1}{2}\) nên A > B