If a sum of $L$ positive integers grows like $L^d$, how does the summand grow?

Question

Suppose that $(a_N)_{N \in \mathbb{N}}$ is a sequence of (strictly) positive integers which satisfies the following property, namely there exists $C \in (0, \infty)$ and an integer $d \in \mathbb{N}$, $d > 1$, such that for any $L \in \mathbb{N}$, $$ \sum\limits_{N=0}^{L} a_N \leq C \, \, L^d. $$ Does this imply that there exists $C^{\prime} \in (0, \infty)$ such that the following inequality is fulfilled for any $N$? $$ a_N \leq C^{\prime} N^{d-1} $$ Of course an obvious bound is $a_N \leq C^{\prime}N^{d}$, but it seems to me that it can be improved.

Answer 1

Clearly the first term does not matter, and we can start summing from $N=1$. Thus you are given information about the sequence of averages of a sequence of positive integers. Namely,

$$\frac{1}{L}\sum_{N=1}^La_N\leq CL^{d-1}$$

and you want to deduce a point-wise growth condition, on the summands. That is probably impossible, because I can imagine infinitely large spikes along the way that do not affect the growth of the average, but do affect of course individual summands. To be concrete, let's start with $a_n=n$. Then clearly the averages grow linearly with $L$. Now let's pick a sequence that has huge gaps, say, $3^n$, and in those places, replace our $a_n$ by $n^2$, say. Then this will not affect the growth of the averages, but will drastically influence the individual elements.