已知正数列$\{a_n\}$对任意自然数$m,n$满足$a_{m+n}\leqslant a_m+a_n$,
证明数列$\left\{\frac{a_n}{n}\right\}$收敛.
\[
0<a_n\leqslant a_{n-1}+a_1\leqslant a_{n-2}+2a_1\leqslant\cdots\leqslant na_1,
\]
\[
\frac{a_{n+1}}{n+1}-\frac{a_{n}}{n}=\frac{na_{n+1}-(n+1)a_n}{n(n+1)}
=\frac{n(a_{n+1}-a_n)-a_n}{n(n+1)}=\frac{na_1-a_n}{n(n+1)}\geqslant 0,
\]
\[
\text{数列}\left\{\frac{a_n}{n}\right\}\text{单调有界, 因此收敛.}
\]
转载于:https://www.cnblogs.com/yinjc/p/4780379.html