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What is $$\lim_{x\rightarrow\,-\infty}\sum_{n=1}^{\infty}\frac{x^n}{n^n}$$?

What is known about specific values of this function at say $-2,-1,0,1,2,3$?

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  • $\begingroup$ For any $x$, the value of the infinite sum is bounded below by $x$. Therefore, as $x$ goes to infinity, so does the sum. EDIT: this reasoning doesn't work if $x$ goes to negative infinity $\endgroup$ – Zubin Mukerjee Aug 2 '17 at 2:31
  • $\begingroup$ @ZubinMukerjee I think it's a bit more subtle, as the limit in question is as $x$ goes to minus infinity $\endgroup$ – TomGrubb Aug 2 '17 at 2:32
  • $\begingroup$ @ThomasGrubb missed the minus sign! thanks $\endgroup$ – Zubin Mukerjee Aug 2 '17 at 2:34
  • $\begingroup$ @ChargeShivers I assume the OP wants to know the values of the series for $x=-2,-1,0,1,2,3$, in addition to asking about the limit. $\endgroup$ – Reveillark Aug 2 '17 at 3:30
  • $\begingroup$ The function defined by a series :$\quad f(x)=\sum_{n=1}^{\infty}x^n/n^n\quad$ cannot be expressed with a finite number of elementary function. A closed form requires the definition of a special function. fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function $\endgroup$ – JJacquelin Aug 2 '17 at 6:50
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In this posting, several users proved that

$$ \lim_{x\to-\infty} \sum_{k=1}^{\infty} \frac{x^k}{k^k} = -1. $$

A slight generalization is also possible: In my previous answer, I proved that

$$ \lim_{x\to-\infty} \sum_{k=1}^{\infty} \frac{x^k}{k^{k-m}} = \begin{cases} -1, & m = 0 \\ 0, & m = 1, 2, \cdots \end{cases} $$


On the other hand, I am not sure if any non-trivial special values for this power series is known. Although I am quite skeptical, one might benefit from the following integral representation

$$ \sum_{k=1}^{\infty} \frac{x^k}{k^k} = \int_{0}^{1} x \cdot u^{-xu} \, du $$

which is a generalization of the Sophomore's dream.

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