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Are the coefficients of power series expansion for a real analytic function bounded? $f(x)=\sum_{n=0}^{\infty}a_n (x-x_0)^n$ We have a sequence $\{a_n\}, n=0,\cdots,\infty$. Is this sequence bounded? Thanks.

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Consider $f(x) = \frac{1}{1 - 2x}$ expanded at $x = 0$. Then $f(x) = \sum 2^n x^n$, so $a_n = 2^n$ is unbounded.

If the radius of convergence is greater than 1, or at least converges at (at least one of) $x_0 \pm 1$, then the terms are guaranteed to be bounded. This is simply because the terms in a convergent sum must go to zero, hence $|a_n| \to 0$, from which you can conclude that $a_n$ is bounded.

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If the radius of convergence is greater than $1$, then the sequence must be bounded, because we would have $$ \sum_{k=0}^{\infty}a_{k} = f(x_{0}+1) < \infty $$ since $x_{0}+1$ lies within the interval of convergence. Since the series of coefficients converges, the sequence of coefficients itself must converge to $0$, which implies that the sequence is bounded.

On the other hand, if the radius of convergence is less than or equal to $1$, the coefficients may be unbounded. Just consider \begin{align*} f(x) = \sum_{k=0}^{\infty}(k+1)x^{k} &= \frac{d}{dx}\left[\sum_{k=0}^{\infty}x^{k}\right] \\ &= \frac{d}{dx}\left[\frac{1}{1-x}\right] \\ &= \frac{1}{(1-x)^{2}}. \end{align*}

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  • $\begingroup$ For this example, is there a explicit function form for f(x)? $\endgroup$ – Sophia Tang Feb 12 '15 at 19:58
  • $\begingroup$ I've edited my answer to include the explicit form. I hope this helps! $\endgroup$ – Jesse Feb 12 '15 at 21:39
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The sequence $\frac{\log|a_n|}{n}$ is bounded, i.e. there exist $C$, $\rho>0$ so that $|a_n| \le C \cdot \rho^n$ for all $n\ge 0$.

Eg: $(2^n \cdot n^3 )$ yes, $(n!)$ no.

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