A series $\sum a_n$ converges if $a_n$ is a nonnegative sequence satisfying $a_{n+2}<\frac{a_{n}+a_{n+1}}{n+2}$

Question

Suppose $\{a_n\}$ is a nonnegative sequence satisfying $a_{n+2}<\cfrac{a_{n}+a_{n+1}}{n+2}$. How can we show that the series $\sum a_n$ converges? It seems that if we use the recursive inequality repeatedly, then we may obtain an inequality of the form $a_n< f(n) (a_1+a_2)$.. But I cannot find a pattern. Any hints?

Answer 1

Set for all $n$, $b_n = \max\{a_{2n},a_{2n + 1}\}$. Then, for all $n$, $$ a_{2n + 2} \leqslant \frac{a_{2n} + a_{2n + 1}}{2n + 2} \leqslant \frac{2b_n}{2n + 2} = \frac{b_n}{n + 1}, $$ and if $n$ is large enough, $\frac{n + 2}{(n + 1)(2n + 3)} \leqslant \frac{1}{n + 1}$ thus, $$ a_{2n + 3} \leqslant \frac{a_{2n + 1} + a_{2n + 2}}{2n + 3} \leqslant \frac{b_n + b_n/(n + 1)}{2n + 3} = \frac{(n + 2)b_n}{(n + 1)(2n + 3)} \leqslant \frac{b_n}{n + 1}. $$ We deduce that for $n \geqslant N$ large enough, $b_{n + 1} \leqslant \frac{b_n}{n + 1}$ so $b_n \leqslant \frac{N!b_N}{n!}$ In particular, the series of the $b_n$ converges, so does the series of the $a_n$.