# Equivalent of the sum $\sum_{n=1}^\infty\frac{x^n}{\sqrt{n}}$

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 $1^-$ ?

• This question is related, and the method proposed in my answer can be applied here. – Antonio Vargas Jan 28 '15 at 22:29
• Such a function is just $\operatorname{Li}_{\frac{1}{2}}(x)$. Near $x=1^-$ its asymptotic behaviour is given by $\zeta\left(\frac{1}{2}\right)+\sqrt{\frac{\pi}{1-x}}.$ – Jack D'Aurizio Jan 28 '15 at 22:37
• @AntonioVargas How would you compute the related integral, then ? – Hippalectryon Jan 28 '15 at 23:17
• I asked Mathematica, and it said $$\int_0^\infty \frac{x^n}{\sqrt{n}}\,dn = \sqrt{\frac{\pi}{-\log x}}$$ :) – Antonio Vargas Jan 28 '15 at 23:43
• @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 :) – Hippalectryon Jan 28 '15 at 23:45

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 $x \to 1^-$.

• 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?) – Hippalectryon Jan 29 '15 at 0:03
• 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. – Antonio Vargas Jan 29 '15 at 0:31
• There's definitely a bit to unwrap in my answer. Trying to prove each $\sim$ yourself would help with understanding. – Antonio Vargas Jan 29 '15 at 3:09