證明對任意正整數 \(n\)，恆有
\[1+\frac{1}{8}+\frac{1}{27}+\frac{1}{64}+\cdots+\frac{1}{n^3}<1.25\].
證明：
先觀常一般項，
\[\frac{1}{n^3} < \frac{1}{n^3-n}=\frac{1}{\left(n-1\right)n\left(n+1\right)} = \frac{1}{2}\left(\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right),\;\forall n>1.\]
所以，
\[1+\frac{1}{8}+\frac{1}{27}+\frac{1}{64}+\cdots+\frac{1}{n^3} < 1+\frac{1}{1\cdot2\cdot3} + \frac{1}{2\cdot3\cdot4}+\cdots+\frac{1}{\left(n-1\right)n\left(n+1\right)}\]
\[=1+\frac{1}{2}\left\{\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)+\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)+\cdots+\left(\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right)\right\} = 1+\frac{1}{4} -\frac{1}{2n\left(n+1\right)}<\frac{5}{4}.\]
故，對任意正整數 \(n\)，\(\displaystyle 1+\frac{1}{8}+\frac{1}{27}+\frac{1}{64}+\cdots+\frac{1}{n^3}<1.25\) 恆成立.
111.7.18補充
95台中一中，
https://math.pro/db/thread-987-1-1.html
