Asymptotic behaviour of $\sum_{i=n}^{\infty} a_i^2$ if $\sum_{i=1}^{\infty} a_i^2 i^{2\gamma} < \infty$ Consider a sequence of positive numbers $(a_i)_{i \in \mathbb{N}}$ such that for some $\gamma > 0$ we have
$$
\sum_{i=1}^{\infty} a_i^2 i^{2\gamma} < \infty.
$$
I was wondering how the tail sum $\sum_{i=n}^{\infty} a_i^2$ behave asymptotically, for $n \to \infty$.
As a heuristic, consider $a_i = i^{-\gamma - \frac{1}{2} - \epsilon}$ for some $\epsilon > 0$. We know this satisfies the assumption and you can show that $\sum_{i=n}^{\infty} a_i^2 \approx n^{-2\gamma - 2\epsilon}$ by comparing it to $\int_n^{\infty} x^{-2\gamma - 1 - 2\epsilon} dx$. So intuitively I would think that for general $(a_i)$, the tail sum can be bounded by $c\ n^{-2\gamma}$ for some $c > 0$.
Just using Cauchy-Schwartz, gives us
$$
\sum_{i = n}^{\infty} a_i^2
= \sum_{i=n}^{\infty} a_i^2 i^{\gamma} i^{-\gamma}
\leq \sqrt{\sum_{i=n}^{\infty} a_i^4 i^{2\gamma} \sum_{i=n}^{\infty} i^{-2\gamma}}.
$$
We know the first sum is finite (since $a_i \to 0$), and the second sum behaves as $\sum_{i=n}^{\infty} i
^{-2\gamma} \approx n^{-2\gamma + 1}$, again using the integral trick. Therefore the tail behaves as $n^{-\gamma + 1/2}$. This doesn't seem to be the best possible.
You could also write
$$
\sum_{i=n}^{\infty} a_i^2
= \sum_{i=n}^{\infty} a_i i^{\gamma} a_i i^{-\gamma}
\leq \sqrt{\sum_{i=n}^{\infty} a_i^2 i^{2\gamma} \sum_{i=n}^{\infty} a_i^2 i^{-2\gamma}}.
$$
Again, the first term is finite, and it seems that the second sum should have the desired asymptotic decay (when considering it with $a_i = i^{-\gamma - \frac{1}{2} - \epsilon}$ for example).
However I don't see how I show this last part in general. What am I missing?

Your tail sum satisfies $$\lim_{n \rightarrow \infty }(\sum_{i=n}^{\infty}a_{i}^2)n^{2\gamma}  = 0$$
To prove this, we can set $a_{i}^2 = b_{i}$ and $2\gamma = \beta$
Note;
$$\sum_{j=n}^{m} b_{j} = \sum_{j=n}^{m}b_{j}j^{\beta}j^{-\beta}. $$
Set $$B(x) = \sum_{0 \leq a \leq x} b_{a}a^{\beta}$$
with $b_{0} = 0$. By Abel Summation formula (https://en.wikipedia.org/wiki/Abel%27s_summation_formula) we have
$$\sum_{j=n}^{m}b_{j}j^{\beta}j^{-\beta}  = B(m)m^{-\beta} - B(n)n^{-\beta} - \int_{n}^{m} B(x)(-\beta n^{-\beta-1})$$
Clearly we can take $m \rightarrow \infty$ to get
$$\sum_{j=n}^{\infty}b_{j} = -B(n)n^{-\beta}+\int_{n}^{\infty} B(x)(\beta n^{-\beta-1})$$
$$=\int_{n}^{\infty} \beta (B(x)-B(n)) n^{-\beta -1}$$
$$= o(n^{-\beta})$$
