Is $\limsup_{t} \frac{a_t}{ h(t+1) - h(t) }$ finite when $0 \leq \sum_{i=t}^\infty a_i \leq h(t) = t^{-\alpha}$? 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.
 A: 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
A: I think the proof gets easier if you replace all the discrete stuff by continuous stuff (eg integrals instead of sums and $h'$ instead of $h(t+1)-h(t)$). I think it is not hard to do some bounds why the continuous problem is equivalent.
If $\frac{a(T)}{h'(T)}=c$, then $a(t)\ge c h'(t)\ \forall t\le T$ because of monotonicity of $a$ and $a(t)\ge 0\ \forall t> T$
For $\alpha=1$ one can show this way that $\limsup_{t\to\infty}\frac{a(t)}{h'(t)}\leq 4$ by solving this optimization problem: https://www.wolframalpha.com/input/?i=min+1%2Ft-%28T-t%29%28T%29%5E%28-2%29*c+from+t%3D0+to+%E2%88%9E (one should optimize $t$ from $0$ to $T$, but it is equiavelnt to optimize from $0$ to $\infty$ which is easier for WolframAlpha) coming from the integral condition $h(t)-\int_t^{\infty}a(s)ds\ge 0$. (You see that the minimum would be negative for $c>4$).
Probably similar argumetns are posssible for $\alpha>1$.
