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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 $a_n=na.$

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up vote 1 down vote accepted

I think it would be better to show more steps in the proof. You could lead up to the first by saying $|a_{2m}-2a_m|\le \frac 1{2m}$, $|a_{3m}-3a_m|\le |a_{3m}-a_{2m}-a_m|+|a_{2m}-2a_m| \le \frac 1{2m}+\frac 1{3m}$ to show what you are thinking. Similarly for the second, you can say $\frac 1{mn} |na_m-ma_n|=\frac 1{mn}|na_m-mna_1|+\frac 1{mn} |mna_1-ma_n|\lt \ldots$ Finally you want to say you can take $n$ and $m$ very large and the right side goes to zero to get that $\frac {a_n}n=\frac {a_m}m$. The approach is fine, but you are making the reader work a bit.

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