Đặt \(A=\frac{1}{2^3}+\frac{1}{3^3}+...+\frac{1}{2019^3}\)
\(\Rightarrow2A=\frac{2}{2^3}+\frac{2}{3^3}+...+\frac{2}{2019^3}\)
Ta có:
\(\left\{{}\begin{matrix}\frac{2}{2^3}< \frac{2}{1.2.3}\\\frac{2}{3^3}< \frac{1}{2.3.4}\\....\\\frac{2}{2019^3}< \frac{2}{\left(2019-1\right).2019.\left(2019+1\right)}\end{matrix}\right.\)
\(\Rightarrow2A< \frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{\left(2019-1\right).2019.\left(2019+1\right)}\)
\(\Rightarrow2A< \frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{\left(2019-1\right).2019}-\frac{1}{2019.\left(2019+1\right)}\)
\(\Rightarrow2A< \frac{1}{1.2}-\frac{1}{2019.\left(2019+1\right)}\)
\(\Rightarrow2A< \frac{1}{1.2}-\frac{1}{2019.2020}\)
\(\Rightarrow A< \left(\frac{1}{1.2}-\frac{1}{4078380}\right):2\)
\(\Rightarrow A< \frac{1}{1.2}:2-\frac{1}{4078380}:2\)
\(\Rightarrow A< \frac{1}{4}-\frac{1}{8156760}\)
\(\Rightarrow A< \frac{1}{2^2}-\frac{1}{8156760}\)
Vì \(\frac{1}{2^2}-\frac{1}{8156760}< \frac{1}{2^2}.\)
\(\Rightarrow A< \frac{1}{2^2}\left(đpcm\right).\)
Chúc bạn học tốt!
