TITLE: A question on a real sequence
QUESTION [14 upvotes]: Let $\{a_n\}_{n\ge1}$ be a real sequence that decays faster than any algebraic speed, that is, $\lim_{n\to \infty} n^pa_n = 0$ for every positive integer $p$. Assume that $$\sum_{n\ge 1}(n+1)^kn^ka_n = 0$$ for every integer $k \ge 0$.
Question: Can we conclude that $a_n \equiv 0$?

REPLY [14 votes]: Counterexample. Consider the analytic function in the unit disk
$$f(z)=\exp\left(-\sqrt{\frac{1}{1-z}}\right)=a_0+a_1z+\ldots,\quad |z|<1,$$
where the principal branch of the $\sqrt{\;}$ is used.
This is the definition of our sequence $a_n$. Function $f$ extends to a $C^\infty$ function on the unit circle, which evident at every point except $z=1$, and it tends to $0=f(1)$ exponentially
as $z\to 1$. So we have that the sequence $(a_n)$ has your property: $|a_n|$ tends to $0$ faster than any
negative power of $n$ (as Fourier coefficients of a $C^\infty$ function). Function $f$ and all derivatives of $f$ vanish at the point $1$. This implies (by Tauber's theorem)
$$\sum_{n=0}^\infty n(n-1)(n-2)\ldots (n-k)a_n=0.$$
for every $k\geq 0$. This easily implies that
$$\sum_{n=0}^\infty n^ka_n=0$$
for every $k\geq 0$, and then
$$\sum_{n=0}^\infty n^k(n+1)^ka_n=0$$
for every $k\geq 0$.