# A question on a real sequence

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 $$a_n \equiv 0$$?

• Define $f(x) := \sum_{n \ge 1} a_nx^{n^2+n}$ for $x \in [0,1]$. Then $f(1),f'(1),f''(1),f'''(1),\dots = 0$. Maybe this is helpful. Jan 22, 2021 at 7:34
• Oh now I see it, you do $f'(1)=0$; $f''(1)=f''(1)+f'(1)=0$ etc Jan 22, 2021 at 8:54
• @PietroMajer well, the 10-th term, for example, tends to infinity. So the maximal term tends to infinity. Jan 22, 2021 at 11:42
• @FedorPetrov: right; see also Eremenko's answer. I was surprised that the problem is stated in an excessively complicated form and wanted to make sure I get the things right.
– Seva
Jan 22, 2021 at 19:25
• Many thanks for your suggestions and thanks to @Alexandre for your counter-example. I understand that $\sum n^ka_n = 0$ for every $k \ge 0$ is not enough to conclude $a_n = 0$. I also understand that $\sum (n+1)^kn^k a_n = 0$ implies $\sum n^k a_n =0$. But how does $\sum n^k a_n = 0$ imply $\sum (n+1)^kn^k a_n = 0$ as mentioned in Alexandre's answer? Jan 23, 2021 at 1:20

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 $$k\geq 0$$.
• for $z=-1$ the smoothness is not evident for me Jan 22, 2021 at 19:14
• @FedorPetrov: Replacing $1+z$ by $2+z$ should fix that (or by $1$, for that matter). Jan 22, 2021 at 22:44