Let $a_n\geqslant 0$ and $s_N = a_0 + \cdots a_{N-1}$, and suppose that $s_N = O\left(N^2\right)$. Then is it inevitable that $a_n=O\left(n\right)$? Looks obvious but not trivial to prove.
Context: in the route to proving that $\sum_{m,n\geqslant 1} \left(mz+n\right)^{-3}$ converges absolutely provided $z\notin\mathbb R$, I want to show that $$\#\left\{N<\left|mn+z\right|\leqslant N+1\right\}=O\left(n\right).$$ I know that this is easy to show with more elementary arguments, but I was curious about the question in the title.
No, it is not true. Let $a_{2^k} = 2^{2k}$, with $a_n = 0$ if $n$ is not a power of $2$. Then it is not hard to show $s_N = O(N^2)$, but $a_n$ is not $O(n)$.