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Question: Find necessary and sufficient condition on the sequence $(a_n)_{n=1}^∞$ so that$$\lim_{n→∞}\sum_{k=0}^{\lfloor\frac1{a_n}\rfloor}(-1)^k\binom nk(1-ka_n)^{n-1}=1\tag 1$$given that $\lim\limits_{n\to\infty}a_n=0$ and $a_n\gt 0$ for all $n\in\Bbb{N}$.

After some guesswork I got to a condition that if $\sum\limits_{n\ge 1} a_n=\infty$ then eq.(1) holds. But I was not able to prove it neither could I find a counterexample for the conjecture. Searching on internet I found that this sum is very closely related to a special case of Dvoretzky covering problem but still couldn't find the necessary and sufficient condition. Until now I have tried using approximations for the Binomial Coefficient and binomial approximation to tackle the sum to no avail. I would be glad if someone could help.

Edit: I have got a counterexample for my conjecture i.e. $\sum\limits_{n\ge 1} a_n=\infty$ is alone not sufficient for eq.(1) to hold. So what should be the necessary and sufficient condition?

Shaurya
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2 Answers2

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This isn't an answer so much as me reporting what I've found simply testing different $a_n$ sequences. First, if $a_n=\frac{1}{n^p}$ ($p\in\mathbb{N}$) then the sum is always zero. Also, if $a_n$ grows as fast or faster than $\frac1n$ then the sum converges to zero. Now, one case which did go to $1$ in the limit was

$$a_n=\frac{\log(n^a)}{n}$$

for $a>1$. Unfortunately, I can't tell what happens when $a=1$ (it may very well converge to $1$) but at $a\in\{2,3,1.5,...\}$ it always seems to converge to $1$.

Again, this is not an answer, but if I were you I would investigate the function $\frac{\log(n^a)}{n}$ and see if that might be some sort of cutoff point.

QC_QAOA
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  • I already did investigate the form $\displaystyle a_n=\frac{1}{n^k}$. For $k\in(0,1)$ The limit is $1$ but for $k\ge 1$ the limit is $0$. For the case of $\displaystyle a_n=\frac{\ln(n^a)}{n}$,if $a=1$ then limit is some constant lying between $0$ and $1$ whereas for the case of $a\in(0,1)$ it does seem to converge to 0 ( but very slowly so I may be wrong too.) – Shaurya Apr 26 '20 at 08:50
  • Interesting, maybe one approach could be to prove that the limit does exist (regardless of $a_n$) and then show something about that limit in terms of $a_n$? – QC_QAOA Apr 26 '20 at 12:46
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somewhat belated answer, but the necessary and sufficient condition for the coverage is the divergence of the series $$\sum_{n\geq1}\frac{1}{n^2}\exp\Big(\sum_{l=1}^n a_l\Big).$$

A reference to the result is "Covering the line with random intervals" by L. A. Shepp, Zeitschrift für Wahrscheinlichkeitstheorie und Verwandte Gebiete volume 23, pages 163–170 (1972).

Peter Strouvelle
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