A sequence $(a_n)$ satisfies: $|a_{m+n}-a_{m}-a_{n}|<\frac{1}{m+n}$ for all positive integers $m,n$ Show that $(a_n)$ is an arithmetic progression.
I thought to solve it like this way :
by induction $|a_{km}-ka_m|\le \frac 1m (\frac 12 +\frac 13 +...+\frac 1k)=\frac{H_k-1}{m}$. Therefore $|\frac{a_m}{m}-\frac{a_n}{n}|=\frac{1}{mn}|na_m-ma_n|<\frac{1}{mn}(\frac{H_n-1}{m}+\frac{H_m-1}{n})$. Therefore $\frac{a_n}{n}=const =a$ and $a_n=na.$