\(E=\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+\cdots+\frac{1}{2021^3}\) Ta có
\(\frac{1}{2^3}=\frac18<\frac16=\frac{1}{1.2.3}\)
Tương tự
\(\frac{1}{3^3}<\frac{1}{2.3.4}\)
\(\frac{1}{4^3}<\frac{1}{3.4.5}\)
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\(\frac{1}{2021^3}<\frac{1}{2020.2021.2022}\)
Do đó
\(\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+\cdots+\frac{1}{2021^3}<\frac{1}{2.3.4}+\frac{1}{3.4.5}+\ldots+\frac{1}{2020.2021.2022}\)
\(\Rightarrow E<\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+\cdots+\frac{1}{2020.2021.2022}\)
\(\Rightarrow2E<\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+\cdots+\frac{2}{2020.2021.2022}\)
\(\Rightarrow2E<\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+\cdots+\frac{1}{2020.2021}-\frac{1}{2021.2022}\)
\(\Rightarrow2B<\frac{1}{1.2}-\frac{1}{2021.2022}\)
\(\Rightarrow E<\frac12\left(\frac{1}{2.3}-\frac{1}{2021.2022}\right)\)
\(\Rightarrow E<\frac{1}{2^2}-\frac{1}{2.2021.2022}<\frac{1}{2^2}\)
\(\Rightarrow E<\frac{1}{2^2}\left(đpcm\right)\)