\(a,\Leftrightarrow2A=\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+...+\dfrac{2}{\left(n-1\right)n\left(n+1\right)}\\ \Leftrightarrow2A=\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+1\right)}\\ \Leftrightarrow2A=\dfrac{1}{2}-\dfrac{1}{n\left(n+1\right)}=\dfrac{n\left(n+1\right)-2}{2n\left(n+1\right)}\\ \Leftrightarrow A=\dfrac{n\left(n+1\right)-2}{4n\left(n+1\right)}\)
\(b,\) Ta cần cm \(\dfrac{1}{k^3}< \dfrac{1}{2}\left[\dfrac{1}{\left(k-1\right)k}-\dfrac{1}{k\left(k+1\right)}\right]\)
\(\Leftrightarrow\dfrac{1}{k^3}< \dfrac{1}{2}\cdot\dfrac{2}{\left(k-1\right)k\left(k+1\right)}\\ \Leftrightarrow\dfrac{1}{k^3}< \dfrac{1}{\left(k-1\right)k\left(k+1\right)}\left(luôn.đúng\right)\)
Do đó \(B< \dfrac{1}{2}\left(\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+\dfrac{1}{3\cdot4}-\dfrac{1}{4\cdot5}+...+\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+1\right)}\right)\\ \Leftrightarrow B< \dfrac{1}{2}\left(\dfrac{1}{6}-\dfrac{1}{n\left(n+1\right)}\right)< \dfrac{1}{2}\cdot\dfrac{1}{6}=\dfrac{1}{12}\)
