$${{3}^{2019}}=a$$，$$A=\frac{a+1}{3a+1}$$，$$B=\frac{3a+1}{9a+1}$$，  $$\frac{1}{A}=\frac{3a+1}{a+1}=\frac{3(a+1)-2}{a+1}=3-\frac{2}{a+1}$$，  $$\frac{1}{B}=\frac{9a+1}{3a+1}=\frac{3(3a+1)-2}{3a+1}=3-\frac{2}{3a+1}$$，  ∴$$\frac{1}{A} ~\textless{} ~\frac{1}{B}$$，  ∴$$A\textgreater B$$． 
Known $$A=\frac{{{3}^{2019}}+1}{{{3}^{2020}}+1}$$，$$B=\frac{{{3}^{2020}}+1}{{{3}^{2021}}+1}$$, compare between $$A$$ and $$B$$． 
Options:
[A] $A=B$ 
[B] $A\textgreater B$ 
[C] $A\textless B$ 
[D] All of the above 
[E] None of the above 