Câu 6:
\(B=\left(\dfrac{1}{199}+1\right)+\left(\dfrac{2}{198}+1\right)+...+\left(\dfrac{198}{2}+1\right)+1\)
\(=\dfrac{200}{199}+\dfrac{200}{198}+...+\dfrac{200}{2}+\dfrac{200}{200}\)
\(=200\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{198}+\dfrac{1}{199}\right)\)
=200A
=>A/B=1/200