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That's it. I want to know a way to do it so i can use it to prove other results like this ones: $$ \lim_{x \to 1^-} \left(\left(\sum_{n=1}^{\infty}x^{n^2} \right)\cdot \sqrt{\log\left(\frac{1}{x}\right)} \right)= \frac{\sqrt{\pi}}{2}$$ $$ \lim_{x \to 1^-} \left(\left(\sum_{n=1}^{\infty}x^{n^m} \right)\cdot \log\left(\frac{1}{x}\right)^{1/m} \right)= \frac{\Gamma \left(\frac{1}{m}\right)}{m}$$

Ramil
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Rafa
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    What if you replace the sum with an integral and show that that's valid in the limit? – Pillsy Apr 09 '17 at 18:37
  • How? I don't really get what you mean, sorry... – Rafa Apr 09 '17 at 18:39
  • The first one ($~$with $x = 1 - \epsilon~$): $\epsilon^{1/2}\int_{0}^{\infty}\exp\left(-\epsilon z^{2}\right),\mathrm{d}z,,,\stackrel{\mathrm{as}\ \epsilon\ \to\ 0^{+}}{\to} {\sqrt{\pi} \over 2}$. The second one follows a similar pattern. – Felix Marin Apr 10 '17 at 02:23

1 Answers1

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Hint. Assume $m>0$ and $0<x<1$. The function $t \mapsto e^{-(-\ln x)\cdot t^m}$ being monotone decreasing over $[1,\infty)$, one deduces $$ \int_1^\infty e^{-(-\ln x)\cdot t^m}\,dt\le\sum_{n=1}^\infty x^{n^m} \le x+\int_1^\infty e^{-(-\ln x)\cdot t^m}\,dt $$ then one may evaluate $$ \int_1^\infty e^{-(-\ln x)\cdot t^m}\,dt. $$ For example, $$ \begin{align} \int_1^\infty e^{-(-\ln x)\cdot t}\,dt&=-\frac{x}{\ln x}, \\\\ \int_1^\infty e^{-(-\ln x)\cdot t^2}\,dt&=\frac{\sqrt{\pi}\text{erfc}\left(\sqrt{-\ln x}\right)}{2\sqrt{-\ln x}}. \end{align} $$

Olivier Oloa
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