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Suppose for each $s$ in an open interval, $P_s(x)=\sum_{k=0}^\infty a_k(s) x^k$ is a power series with radius of convergence greater than R, where each $a_k(s)$ is differentiable. My question is: Is the radius of convergence of $\sum\limits_{k=0}^\infty a_k'(s) x^k$ at least R?

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The radius of convergence can be $0$ for some $s$. Let $h \colon \mathbb{R}\to\mathbb{R}$ a continuous (or smooth, if you want your coefficients smooth) function with support in $[-1,1]$, $h(0) = 1$, $h \geqslant 0$ and $\int_{-1}^1 h(t)\,dt = 1$. Let

$$a_k(s) = k!\int_0^s h((k!)^2\cdot t)\,dt.$$

Then you have $\lvert a_k(s)\rvert \leqslant \frac{1}{k!}$, so the radius of convergence is $\infty$ for all $s$, but $a_k'(0) = k!$, so

$$\sum_{k=0}^\infty a_k'(0)x^k$$

has radius of convergence $0$.

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  • $\begingroup$ Is it possible if a_k's are analytic? $\endgroup$ – TCL May 9 '14 at 17:33
  • $\begingroup$ You can take $a_k(s) = (k!s) e^{-(k!s)^2}$, then the $a_k(s)$ are uniformly bounded, so the radius of convergence is at least $1$ for all $s$, but $a_k'(0)= k!$. If $f(s,x) = \sum a_k(s)\cdot x^k$ is analytic in both variables jointly, then it cannot happen. $\endgroup$ – Daniel Fischer May 9 '14 at 17:43
  • $\begingroup$ Perhaps your last comment here answer my other question? since the coefficents A_i there are derivatives of rational functions in s. If so, can you answer the question there?:math.stackexchange.com/questions/787379/… $\endgroup$ – TCL May 9 '14 at 21:58

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