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 $a\_n \equiv 0$?



A question on a real sequence
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 $k\geq 0$.


