TITLE: Is $\limsup_{t} \frac{a_t}{ h(t+1) - h(t) }$ finite when $0 \leq \sum_{i=t}^\infty a_i \leq h(t) = t^{-\alpha}$?
QUESTION [6 upvotes]: Consider a monotone decreasing sequence $\{a_i\}_{i=1}^\infty$, where $a_i  \geq 0$ for all $i$. This sequence is assumed to satisfy the bound on its residual series
$$
\sum_{i=t}^\infty a_i \leq h(t) = t^{-\alpha} \;\;\;\forall t\in\mathbb{N}
\;,
$$
where $\alpha \geq 1$. From the above bound, it is clear that $\{a_i\}$ is summable since we have $\sum_{i=1}^\infty a_i \leq 1 < \infty$.
My objective is to determine whether
$$\limsup_{i\rightarrow \infty} \frac{a_i}{-\Delta h_i} < \infty \,,$$
where $\Delta h_i = h(i+1) - h(i)$.
Thus far, I have only been able to prove the weaker property that
$$ \liminf_{i\rightarrow\infty}  \frac{a_i}{-\Delta h_i} \leq 1 \;,$$
which is done by showing that $a_i \leq -\Delta h_i$
is true for infinitely many $i$ (see below for proof).
I have found that the answer may be related to the Stolz-Cesaro Theorem or its converse, but I have been unable to make the exact requirements of the theorem line up with what I need. I have also found that this may be connected to the following post.
Any help on this would be appreciated. I am also curious about the more general case when we have some convex function $h:(0,\infty)\rightarrow (0,\infty)$ satisfying $\lim_{i\rightarrow\infty} h(i) = 0$.

Proof that $a_i \leq -\Delta h_i$ for infinitely many $i$: Assume the converse for all but finitely many $i$. Then there exists a $T\in\mathbb{N}$ such that for all $t \geq T$,
$$
\sum_{i=t}^\infty a_i > - \sum_{i=t}^\infty \Delta h_i = h(t)
\;,
$$
which is a contradiction. Hence we must have that $a_i \leq -\Delta h_i$ holds infinitely often.

REPLY [4 votes]: You see for all $t$,by monotonicity of $a$,  we have:
$$ t^{\alpha+1} a_{2t}  \le t^{\alpha} \sum_{n \ge t+1} a_n \le 1$$
Hence, $$ \limsup t^{\alpha+1} a_{t} \le 2^{\alpha+1}$$
Q.E.D