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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 $a_n \equiv 0$?

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    $\begingroup$ 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. $\endgroup$ Jan 22 at 7:34
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    $\begingroup$ Oh now I see it, you do $f'(1)=0$; $f''(1)=f''(1)+f'(1)=0$ etc $\endgroup$ Jan 22 at 8:54
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    $\begingroup$ @PietroMajer well, the 10-th term, for example, tends to infinity. So the maximal term tends to infinity. $\endgroup$ Jan 22 at 11:42
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    $\begingroup$ @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. $\endgroup$
    – Seva
    Jan 22 at 19:25
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    $\begingroup$ 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? $\endgroup$
    – Jacob Lu
    Jan 23 at 1:20
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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 $k\geq 0$.

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    $\begingroup$ for $z=-1$ the smoothness is not evident for me $\endgroup$ Jan 22 at 19:14
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    $\begingroup$ @FedorPetrov: Replacing $1+z$ by $2+z$ should fix that (or by $1$, for that matter). $\endgroup$ Jan 22 at 22:44
  • $\begingroup$ @Christian Remling: thanks for your suggestion. $\endgroup$ Jan 23 at 0:39

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