Let $\{b\_n\}\_{n\geq0}$ be a sequence such that $b\_nb\_{n+1}=0$ and define
$$a\_n:=\sum\_{k=0}^n(-1)^{n-k}\binom{n}{k}b\_k.$$
If $\lim\_{n\to\infty}a\_n=0$, can we conclude that $b\_n=0$ for all $n$?


More generally, if $\{b\_n\}\_{n\geq0}$ is a sequence with infinitely many zeros and $\lim\_{n\to\infty}a\_n=0$, can we still conclude that $b\_n=0$ for all $n$?


Finally, what if we only assume $\{b\_n\}\_{n\geq0}$ contains at least ONE zero?


Till now, only the original question remains unsolved.


**Remark:** This question arised from the computation of the $K\_0$ groups of the smooth noncommutative $\mathbb{R}^{2n}$ which come from noncommutative field theory, see [enter link description here](https://arxiv.org/abs/2208.06253). I showed that the $K\_0$ groups of original noncommutative $\mathbb{R}^{2n}$ are all $\mathbb{Z}$. Then mimicking the smooth noncommutative tori, I construct the smooth noncommutative $\mathbb{R}^{2n}$ and I want to show that the $K\_0$ groups of the smooth cases are still all $\mathbb{Z}$. But this brings many new problems. First I consider a special class of the projectors of smooth noncommutative $\mathbb{R}^{2}$, and show that if this conjecture is true, then this class of projectors are 0 or 1. The other two questions I asked here are also related to the characterization of projectors of smooth noncommutative $\mathbb{R}^{2n}$.



A number sequence problem involving binomial transform
Unfortunately the argument that I originally posted contained a gap at the end.
The gap is explained at the end of the proof, where I also state 3 partial results.


Let
$$f(z)=\sum\_{n=0}^\infty b\_nz\_n/n!,\quad g(z)=\sum\_{n=0}^\infty a\_nz^n/n!.$$
Then your relation between $a\_n$ and $b\_n$ means
$$g(z)=e^{-z}f(z).$$ Suppose that $a\_n=o(1)$, then $|g(z)|=o(e^{|z|})$. Thus both $f$ and $g$ are of exponential type, and the indicator diagram $K$ of $f$ is contained in the
disk $D=\{ z:|z+1|\leq 1\}.$ This means that the Borel (=Laplace) transform
$$F(z)=\sum\_{n=0}^\infty b\_n z^{-n-1}$$
is analytic outside of this disk $D$. Thus the power series
$$F(1/z)=\sum\_{n=0}^\infty b\_nz^{n+1}$$
has radius of convergence at least $1/2$ and has an analytic continuation into the half-plane $\{ z:\Re z<1/2\}$. But this contradicts the Fabry Gap Theorem, since
your condition that $b\_nb\_{n+1}=0$ implies that the density
of non-zero coefficients is at most $1/2$, and Fabry's theorem says that under this condition $F(1/z)$ must have
a singularity on every arc of the circle of convergence which is bigger than semi-circle.


This argument does not work when $F(1/z)$ has infinite radius of convergence.


1. To exclude this case, one may assume that $|a\_n|$ tends to zero with geometric speed, that is $|a\_n|=O(\delta^n)$
for some $\delta\in(0,1)$. Then $0\not\in K$, and
$F(1/z)$ is not entire.
2. And of course, if the condition $b\_nb\_{n+1}=0$
is replaced by the condition that $f$ is even or odd, then the conclusion is true as well.
3. The argument above also shows that under the stated conditions one can conclude that $b\_n=O(\epsilon^n),\forall \epsilon>0$.


Thus my argument needs either a stronger assumption
(1) or (2), or leads to a weaker conclusion (3).


Reference.


L. Bieberbach, Analytische Fortsetzung, Springer 1955, Chap. 2
(Chap. 1 also contains all other facts used in this argument).


