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I have mainly two questions, the first one being motivated by the second one.

1) Is there a way to prove that $F(x) = \sum_{k=1}^\infty \frac{(-1)^{k+1} x^{2k}}{(2k)!}$ is bounded on $\mathbb{R}_+$ without using the fact that $F(x) = 1-cos x$?

2) If $(u_k)_{k\in\mathbb{N}}$ is a positive sequence decreasing to $0$, such that $u_k^{1/(2k)}$ converges to $u\in (0,1)$, can I prove that $G(x) = \sum_{k=1}^\infty \frac{(-1)^{k+1} u_k\ x^{2k}}{(2k)!}$ is bounded on $\mathbb{R}_+$?

I have tried several things, in particular comparing the sums of these two series, Abel transform, ... Maybe by adding some restriction on $u_k$ (of the form $\frac{u^{2k-2}}{2k}$ would be enough for my purpose)?

Thanks.

Remark: a more general (unanswered) question was raised in: Criteria for boundedness of power series, and the sequence in 2) matches the necesary condition mentioned in this post.

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    $\begingroup$ This looks hopeless. I don't think we expect this property (boundedness) to have any stability properties of this kind. For a trivial counterexample, let $u_k=2^{-2k}$ for $k\ge 2$, $u_1\not= 2^{-2}$. $\endgroup$ Jun 22, 2014 at 15:48
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    $\begingroup$ Adding to Christian's comment: since the sum of the terms with even $k$ is unbounded (approx $e^x/4$) one can adjust the $u_k$ values by a tiny bit to favour even $k$ and the resulting series will be unbounded. $\endgroup$ Jun 22, 2014 at 20:17
  • $\begingroup$ I think that the most fair approach, pretending not to know the sum of the series, is making some derivatives and discover that it solves a suitable linear differential equation, whence it is periodic. $\endgroup$ Jul 4, 2014 at 12:33

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