A=1/5^3+1/6^3+...+1/2023^3
1/5^3<1/4*5*6
Xét tương tự, ta đều sẽ được:
\(\dfrac{1}{n^3}< \dfrac{1}{n\left(n-1\right)\left(n+1\right)}\)
=>\(A< \dfrac{1}{4\cdot5\cdot6}+\dfrac{1}{5\cdot6\cdot7}+...+\dfrac{1}{2022\cdot2023\cdot2024}\)
=>\(A< \dfrac{1}{2}\left(\dfrac{2}{4\cdot5\cdot6}+\dfrac{2}{5\cdot6\cdot7}+...+\dfrac{2}{2022\cdot2023\cdot2024}\right)\)
=>\(A< \dfrac{1}{2}\left(\dfrac{1}{4\cdot5}-\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}-\dfrac{1}{7\cdot8}+...+\dfrac{1}{2022\cdot2023}-\dfrac{1}{2023\cdot2024}\right)\)
=>A<1/40
Ta có BĐT: \(\dfrac{1}{k\left(k+1\right)\left(k+2\right)}< \dfrac{1}{k^3}< \dfrac{1}{\left(k-1\right)\cdot k\cdot\left(k+1\right)}\)
Do đó, ta được:
\(\dfrac{1}{5\cdot6\cdot7}+\dfrac{1}{6\cdot7\cdot8}+...+\dfrac{1}{2023\cdot2024\cdot2025}< A\)
\(\Leftrightarrow A>\dfrac{1}{2}\left(\dfrac{1}{5\cdot6}-\dfrac{1}{2024\cdot2025}\right)>\dfrac{1}{2}\left(\dfrac{1}{30}-\dfrac{1}{390}\right)=\dfrac{1}{65}\)
=>1/65<A<1/40
