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As you can see here, In Daniel Fischer's answer it says and I quote: "Show that if there is an $m$ such that $a_m\neq 0$, then $|h(1/n)|\leq 1/n!$ cannot hold for large enough $n$".

My question is this: why is this so obvious? I can understand that it would be obvious if $(a_k)\subset [0,+\infty)$, or if only finite terms were non-zero, but in the general case I can't see it. Any help is appreciated.

Intuitively, it seems a contradiction indeed, since LHS is polynomial-like and RHS is factorial, but I can't work out the details. For anyone attempting a proof: keep in mind that the power-series converges absolutely and locally uniformly on the disk $D(0,10)$ (I guess it helps).

Just dropped in
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2 Answers2

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Write $$h(z) = \sum_{k = 0}^{\infty} a_k z^k.$$We have $h(0)=\lim_nh(1/n)=0$. So $a_0=0$. Since the radius of convergence is $10$, we know for instance that $$\sum_k 9^k|a_k|<\infty.$$In particular $9^k|a_k|\to0$, and there exists $c>0$ with $|a_k|<c/9^k$ for all $k$.

If $a_1\ne0$, there exists $n_0$ such that $\frac{c}{81n^2-9n}\leq\frac{|a_1|}{2n}$ for all $n\geq n_0$. For such $n$, \begin{align} \frac1{n!}&\geq\left|h\left(\frac1n\right)\right|=\left|\frac{a_1}n+\sum_{k=2}^\infty \frac{a_k}{n^k}\right|\geq\frac{|a_1|}{n}-\sum_{k=2}^\infty \frac{|a_k|}{n^k}\geq\frac{|a_1|}{n}-\sum_{k=2}^\infty \frac{c}{(9n)^k}\\ \ \\ &=\frac{|a_1|}{n}-c\,\frac{1/(9n)^2}{1-1/9n} =\frac{|a_1|}{n}-c\,\frac{1}{81n^2-9n}\\ \ \\ &\geq\frac{|a_1|}{2n} \end{align} The above inequality is impossible for $n$ big enough. It follows that $a_1=0$. Now we can use a very similar argument for $a_2$, then for $a_3$, etc., showing that $a_k=0$ for all $k$.

Martin Argerami
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If $h(0)\ne 0,$ then $h(1/n)$ has a nonzero limit at $0,$ hence cannot satisfy the inequality for large $n.$ If $h(0)=0,$ then the order of the zero of $h$ at $0$ is finite (since $h$ is nonconstant). Thus there exists $m>0$ such that $h(z) =z^mg(z),$ where $g$ is analytic near $0$ and $g(0)\ne 0.$ This implies $|h(1/n)|$ is on the order of $1/n^m$ as $n\to \infty.$ Since $1/n!\to 0$ much faster than that, we see the inequality in question cannot hold for large $n.$

zhw.
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