6

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.

Hotdog2000
  • 309
  • From reading your edits I think you mean $-\Delta h_i\geq -\Delta h_{i+1}>0$ I believe you want h to decrease but for the differences to decrease in size. Otherwise $h(t)=t^{-1}$ won't work. – Ciarán Ó Raghaillaigh Dec 08 '20 at 09:19
  • @Ciaran, thank you, I've corrected the direction of the inequality – Hotdog2000 Dec 08 '20 at 13:50
  • For $h(t) = t^{-\alpha}$ there's hope, but in general the quotient can be unbounded (I'm pretty sure). Write $b_i = h(i) - h(i+1)$. Let $(n_k){k \in \mathbb{N}}$ be a sufficiently fast increasing sequence. For $n{k-1} < i < n_{k}$ let $a_i = c_k$ and $b_i = d_k$ with $d_k/c_k \to +\infty$ and $c_k/d_{k+1} \to +\infty$. Also let $a_{n_k} = c_{k}$ and $b_{n_k} = d_{k+1}$. Then $\frac{a_i}{b_i}$ is large for $i = n_k$, but one can (unless I'm mistaken) make the choices to still have $\sum_{i \geqslant t} a_i \leqslant \sum_{i \geqslant t} b_i$. – Daniel Fischer Dec 08 '20 at 15:03
  • @DanielFischer thanks for the comment. I think I may reformulate the problem once more so that $h(t)=t^{-\alpha}$. I think that $h(t)=e^{-\beta t}$ can be shown to work by applying the converse of the Stolz-Cesaro theorem (linked in the post) since $\lim_{t} \frac{h(t)}{h(t+1)} \neq 1$. – Hotdog2000 Dec 08 '20 at 15:33

2 Answers2

4

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

Paresseux Nguyen
  • 7,331
1

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 $\alpha>1$.

Jakob
  • 183