Finding conditions for $\lim\limits_{n\to\infty}\sum\limits_{k=0}^{\lfloor\frac1{a_n}\rfloor}(-1)^k\binom nk(1-ka_n)^{n-1}=1$

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?

• Commented May 21, 2022 at 11:52

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.

• 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.) Commented Apr 26, 2020 at 8:50
• 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 $a_n$? Commented Apr 26, 2020 at 12:46

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).