第二題
先證明遞減 \(a_{n}-a_{n-1}=\frac{a_{n-1}-a_{n-2}}{2}+\frac{a_{n-2}-a_{n-1}}{a_{n-1}a_{n-2}}=(a_{n-1}-a_{n-2})(\frac{1}{2}-\frac{1}{a_{n-1}a_{n-2}})=(a_{n-2}-a_{n-3})(\frac{1}{2}-\frac{1}{a_{n-1}a_{n-2}})(\frac{1}{2}-\frac{1}{a_{n-2}a_{n-3}})\)
\(=(a_{2}-a_{1})(\frac{1}{2}-\frac{1}{a_{n-1}a_{n-2}})(\frac{1}{2}-\frac{1}{a_{n-2}a_{n-3}})\times\cdots \times(\frac{1}{2}-\frac{1}{a_{2}a_{1}})\leq 0\)
(因為\(a_{n+1}a_{n}\geq 2 \forall n\in N \))
\(a_{n}\)就直接帶進去個3項左右可拿到 \(a_{3}=\frac{577}{408}\doteq 1.41422\)
\(a_{3}\leq \sqrt{2}+\frac{1}{2021}\doteq 1.41470\)
最後因為遞減 所以\(a_{3}>a_{2021}\)
第三題的想法可以參考
https://frankliou.wordpress.com/ ... %E6%A5%B5%E9%99%90/
