Maximum of a sequence and convergence of a series

Question

Consider a real sequence $(a_i)_{i \in \mathbb N}$ with $$0\leq a_i \\ \sum_{i=1}^Na_i \leq N$$ Suppose $$\lim_{n\rightarrow \infty}\frac{1}{n}\max_{1 \leq i \leq n}a_i=0$$ Then is it true that

$$\lim_{N\rightarrow \infty}\frac{1}{N^2}\sum_{i=1}^Na_i^2=0$$

Thanks and regards.

Answer 1

I believe the hypothesis $\sum_{i=1}^{N}a_i =N$ should be $\sum_{i=1}^{N}a_i \leq N$, since. otherwise, $a_n=1$ for all $n$. Taking this as hypothesis note that $\sum_{i=1}^{N} a_i^{2} \leq \max \{a_i:1\leq i\leq N\} \sum_{i=1}^{N} a_i$ (because $a_k^{2} \leq \max \{a_i:1\leq i\leq N\} a_k$ for each $k$). Divide by $N^{2}$ and let $N \to \infty$. Since $\frac 1 {N^{2}} \max \{a_i:1\leq i\leq N\} \sum_{i=1}^{N} a_i \leq \frac 1 N \max \{a_i:1\leq i\leq N\} \to 0$ the proof is complete.