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Motivation: I've been thinking about the transformation of power series, which takes the (power series of) $\exp(x)$ to $\sin(x)$. At first i was trying the series $\sum_{n=0}^{\infty} \frac{x^n}{n^n}$. Since for positive $x$ this function is smaller than $\exp(x)$, i was expecting that after the transformation i get a function similar to $\sin(x)$ but with smaller waves. What i got was the exact opposite, the waves were growing in width and in height too. (As far as mathematica could do the numerical celculations in reasonable time.) When i first told this to my calculus teacher, she replied, that i just changed the $n!$ to $n^n$ in the power series of $\sin(x)$. So i'm interested in this function: $$ \sum_{n=0}^{\infty} \frac{(-1)^nx^{2n+1}}{{(2n+1)}^{2n+1}} $$

Question: is this function bounded on the positive reals? Also this question sounds so natural, are there any results concerning this function in the literature?

Daniel Soltész
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    Seems similar to this from last year, on $\sum_{n=0}^\infty x^n/n^n$ for $x \rightarrow -\infty$: http://mathoverflow.net/questions/109160 The same technique should apply here. –  Aug 06 '13 at 22:25
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    ... and, indeed, these are natural questions, and should not be discouraged, but just put in a different venue. (And we oughtn't be surprised that people who reasonably come up with such questions are comparably naive about prior sources/literature, so they'll recur endlessly as new people grow into the relevant state.) – paul garrett Aug 06 '13 at 22:32
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    Well the $\sum_{n=0}^\infty x^n/n^n$ question was not entirely trivial (and it seems that the technique isn't all that well-known). Here we end up with $x \int_0^\infty te^{-t} \cos(xte^{-t}) dt$, which I expect will be dominated by a stationary phase around $t=1$ that will oscillate like $\cos(x/e)$ and reduce the growth from $x$ to $x^{1/2}$. –  Aug 06 '13 at 22:57
  • We could debate whether the question itself belongs here. I have to agree that the presentation of it strongly suggested otherwise. What is more important than the place of the question is that i got a very usefull comment, from Noam D. Elkies. Now i can work on my own for a while. Thank you! – Daniel Soltész Aug 07 '13 at 06:05

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