How to control the tail behavior of a class of series

Assume $a_n$ is a monotonically decreasing non-negative series ($1\geq a_1\geq a_2 \geq ... \geq 0$). And we know it decreases faster than $n^{-2}$ (but potentially it can decrease very fast, or converge to zero after finite terms).

I wonder is it true that $$ \left(\sum_{n:a_n \leq \epsilon} a_n^2 \right) \cdot \epsilon^{(-3/2)} \text{stays bounded as } \epsilon \rightarrow 0 $$ for all $a_n$ that satisfies the above conditions.

Intuition: $a_n = n^{-2}$ is the special one. We can calculate that under this condition $\sum_{a_n \leq \epsilon} a_n^2 \sim \epsilon^{3/2}$. And in general when $a_n = n^{-2s}$, $\sum_{a_n \leq \epsilon} a_n^2 \sim \epsilon^{\frac{4s-1}{2s}}$. When $a_n = \exp(-n)$, we get $\sum_{a_n \leq \epsilon} a_n^2 \sim \epsilon^{2}$. It looks like when $a_n$ decreases faster, the series has a better dependency on $\epsilon$.

Question: My difficulty is coming up with a formal proof, it feels like there is some monotonicity when $a_n$ decreases faster. Are there any tools that you all are aware of that works without assuming $a_n$ takes any specific geometrical/exponential forms? Thank you!

Solution 1:

Welcome to MSE!

All we need to know is that $0 \leq a_n \leq n^{-2}$. Let's let $0 < \epsilon < 1$, since we're taking $\epsilon \to 0$ anyways, and choose $N = \left \lceil \epsilon^{-1/2} \right \rceil$. Moreover, let's let $n_0$ be the first $n$ so that $a_{n_0} < \epsilon$.

Then we have

$$ \begin{aligned} \left ( \sum_{a_n < \epsilon} a_n^2 \right ) \epsilon^{-3/2} &= \left ( \sum_{n=n_0}^\infty a_n^2 \right ) \epsilon^{-3/2} \\ &= \left ( \sum_{n=n_0}^{N+1} a_n^2 + \sum_{n=N+1}^\infty a_n^2 \right ) \epsilon^{-3/2} \\ &\leq \left ( (N+1-n_0) \epsilon^2 + \sum_{n=N+1}^\infty n^{-4} \right ) \epsilon^{-3/2} \\ &\overset{(1)}{\leq} \left ( 2N \epsilon^2 + \int_{N}^\infty x^{-4} \ dx \right ) \epsilon^{-3/2} \\ &\leq \left ( 2N \epsilon^2 + N^{-3} \right ) \epsilon^{-3/2} \\ &\overset{(2)}{\leq} \left ( 2(1 + \epsilon^{-1/2}) \epsilon^2 + (\epsilon^{-1/2})^{-3} \right ) \epsilon^{-3/2} \\ &\leq \left ( 2\epsilon^2 + 2\epsilon^{3/2} + \epsilon^{3/2} \right ) \epsilon^{-3/2} \\ &= 3 + 2\sqrt{\epsilon} \\ &\leq 5 \end{aligned} $$

In step $(1)$ we've used the fact that $\sum_{N+1}^\infty f(n) \leq \int_N^\infty f(x) \ dx$ and the (crummy) bound $N+1 \leq 2N$. In step $(2)$ we've replaced an $N$ in the numerator by $1 + \epsilon^{-1/2}$ and an $N$ in the denominator by $\epsilon^{-1/2}$, since $\epsilon^{-1/2} < N < 1 + \epsilon^{-1/2}$. Lastly, since we're only considering $\epsilon < 1$, we can get a uniform bound.

I hope this helps ^_^