MT【156】特例$a_n=\dfrac{6}{\pi n^2}$

设无穷非负数列$\{a_n\}$满足$a_n+a_{n+2}\ge2 a_{n+1},\sum\limits_{i=1}^{n}{a_i}\le1$,证明:
$0\le a_n-a_{n+1}\le\dfrac{2}{n(n+1)}$

证明：由题意$0\le a_n\le\sum\limits_{i=1}^{n}{a_i}\le1$又由于$1\ge a_{m}-a_n\ge(m-n)(a_{n+1}-a_n)$
(凸数列性质,有定义易得)
故当$m>n$时,$a_{n+1}-a_n\le \dfrac{1}{m-n}\longrightarrow 0 (m\longrightarrow \infty)$
右边由$a_{i}\ge a_n+(i-n)(a_{n+1}-a_n)=a_{n+1}+(n+1-i)(a_{n}-a_{n+1})\ge(n+1-i)(a_{n}-a_{n+1})$
故$1\ge\sum\limits_{i=1}^{n}{a_i}\ge\sum\limits_{i=1}^{n}{(n+1-i)(a_{n}-a_{n+1})}=\dfrac{1}{2}n(n+1)(a_n-a_{n+1})$
即得$0\le a_n-a_{n+1}\le<\dfrac{2}{n(n+1)}$
注:此类数列特例$a_n=\dfrac{6}{\pi n^2}$