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I am able to show that $\lim\limits_{q \rightarrow 1} (1-q)^3 \sum_{n>0} n q^n = 0 $ and $ \lim\limits_{q \rightarrow 1} (1-q)^3 \sum_{n>0} n^2 q^n = 2$ and now I want to prove that

$$ \lim\limits_{q \rightarrow 1} (1-q)^3 \sum_{n>0} n^{1+\epsilon} q^n = 0 $$

for $\epsilon \in (0,1)$. Do you guys have any hints for me doing so? I can't use the methods which I used to proof the first two statements because they are just valid for integral exponents of $n$.

Thanks in advance!

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  • $\begingroup$ Mathematica 9 produces 0 by $$In[1]:=Assuming[epsilon > 0, Limit[PolyLog[-1 - epsilon, q]*(1 - q)^3, q -> 1]]$$. $\endgroup$ – user64494 Jun 24 '13 at 15:27
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If we recognize that the series in question is a polylogarithm,

$$ \newcommand{\Li}{\operatorname{Li}} \sum_{n=1}^{\infty} n^\alpha q^n = \Li_{-\alpha}(q), $$

then we can take advantage of the alternate representation (specifically the second equation under #2 in the link)

$$ \Li_{-\alpha}(q) = \Gamma(1+\alpha) (-\log q)^{-\alpha-1} + \sum_{n=0}^\infty \frac{\zeta(-\alpha-n)}{n!} \,(\log q)^n $$

which holds for $|\log q| < 2\pi$ and $\alpha \neq -1,-2,-3,\ldots$. As $q \to 1^-$ the series on the right converges to $\zeta(-\alpha)$ and the term on the left blows up (if $\alpha > -1$). Since $\log q \sim q-1$ as $q \to 1$ we may conclude that

$$ \lim_{q \to 1^-} (1-q)^{\alpha+1}\Li_{-\alpha}(q) = \Gamma(1+\alpha) $$

for all $\alpha > -1$.

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    $\begingroup$ @user64494 how so? $\endgroup$ – Antonio Vargas Jun 24 '13 at 15:32
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    $\begingroup$ @user64494 I have plotted the expression on my own and I believe you have misread my answer. Note that there is an $\alpha+1$ in the exponent in the limit, not $3$. Mathematica verifies my result: Manipulate[Plot[{(1-q)^(a+1) PolyLog[-a,q],Gamma[1+a]},{q,1/2,1},AxesOrigin->{1/2,0}],{a,-1+1/2,3}] $\endgroup$ – Antonio Vargas Jun 24 '13 at 15:38
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    $\begingroup$ @user64494 yes, and the limit written in my answer provides all that is needed to find that limit: if $\alpha + 1 < 3$ then clearly $$\lim_{q \to 1^-} (1-q)^{3}\Li_{-\alpha}(q) = 0,$$ and if $\alpha + 1 > 3$ then $$\lim_{q \to 1^-} (1-q)^{3}\Li_{-\alpha}(q) = \infty.$$ $\endgroup$ – Antonio Vargas Jun 24 '13 at 15:56
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    $\begingroup$ @henrik691 I would check the two references given on wikipedia for the integral representation using the Hankel contour: Whittaker & Watson 1927, § 12.22, § 13.13 and Gradshteyn & Ryzhik 1980, § 9.553. Using this it's apparently possible to derive the series I used; the steps are outlined on wikipedia in the section I linked to in the answer. $\endgroup$ – Antonio Vargas Jun 24 '13 at 16:08
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    $\begingroup$ Just in case someone is interested: A reference for the alternative representation is arxiv.org/pdf/math/0701743.pdf (Cor 1.2) or people.mpim-bonn.mpg.de/zagier/files/exp-math-1/fulltext.pdf . $\endgroup$ – henrik691 Jun 24 '13 at 19:28

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