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Let's consider $\displaystyle f(x)=\sum_{n=1}^\infty\frac{x^n}{\sqrt{n}}$.

Where $f$ is defined, can we find a closed form for $f(x)$ ?

What would be an equivalent of $f$ near $1^-$ ?

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    $\begingroup$ This question is related, and the method proposed in my answer can be applied here. $\endgroup$ – Antonio Vargas Jan 28 '15 at 22:29
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    $\begingroup$ Such a function is just $\operatorname{Li}_{\frac{1}{2}}(x)$. Near $x=1^-$ its asymptotic behaviour is given by $\zeta\left(\frac{1}{2}\right)+\sqrt{\frac{\pi}{1-x}}.$ $\endgroup$ – Jack D'Aurizio Jan 28 '15 at 22:37
  • $\begingroup$ @AntonioVargas How would you compute the related integral, then ? $\endgroup$ – Hippalectryon Jan 28 '15 at 23:17
  • $\begingroup$ I asked Mathematica, and it said $$\int_0^\infty \frac{x^n}{\sqrt{n}}\,dn = \sqrt{\frac{\pi}{-\log x}}$$ :) $\endgroup$ – Antonio Vargas Jan 28 '15 at 23:43
  • $\begingroup$ @AntonioVargas I don't know how to show that result, but I'd love it if you could show me how if you have some time to write an answer :) $\endgroup$ – Hippalectryon Jan 28 '15 at 23:45
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If we write $x^n = \exp(n\log x)$, then set $u = -n\log x$, the integral mentioned in the comments becomes

$$ \int_0^\infty \frac{x^n}{\sqrt{n}}\,dn = \frac{1}{\sqrt{-\log x}} \int_0^\infty \frac{e^{-u}}{\sqrt{u}}\,du = \frac{\Gamma(1/2)}{\sqrt{-\log x}} = \sqrt{\frac{\pi}{-\log x}}. $$

Then since $-\log x \sim 1-x$ as $x \to 1$ and the terms of the sum are strictly decreasing for $0 < x < 1$, we may conclude that

$$ \sum_{n=1}^{\infty} \frac{x^n}{\sqrt{n}} \sim \int_0^\infty \frac{x^n}{\sqrt{n}}\,dn = \sqrt{\frac{\pi}{-\log x}} \sim \sqrt{\frac{\pi}{1-x}} $$

as $x \to 1^-$.

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  • $\begingroup$ Your answer seems fine, but do you know why Jack D'Aurizio has $\zeta\left(\frac{1}{2}\right)+\sqrt{\frac{\pi}{1-x}}$ as an equivalent ? (Btw the sum starts at $1$ and not $0$, shouldn't the integral start at $1$ too?) $\endgroup$ – Hippalectryon Jan 29 '15 at 0:03
  • $\begingroup$ The term $\zeta(1/2)$ is the "next" term in the approximation and is small compared to the $\sqrt{\cdots}$ term, which is dominant. Both $\sqrt{\cdots}$ and $\sqrt{\cdots} + \zeta(1/2)$ are equivalents to your function, since equivalence is only determined by leading-order behavior. The integral starts at $0$ for simplicity; the difference accrued by it starting at $1$ would be small compared to the dominant contribution of the tail. $\endgroup$ – Antonio Vargas Jan 29 '15 at 0:31
  • $\begingroup$ There's definitely a bit to unwrap in my answer. Trying to prove each $\sim$ yourself would help with understanding. $\endgroup$ – Antonio Vargas Jan 29 '15 at 3:09

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