Limit of $\sqrt{\frac{\pi}{1-x}}-\sum\limits_{k=1}^\infty\frac{x^k}{\sqrt{k}}$ when $x\to 1^-$?

I am trying to understand if

$$\sqrt{\frac{2\pi}{1-x}}-\sum\limits_{k=1}^\infty\frac{x^k}{\sqrt{k}}$$ is convergent for $x\to 1^-$. Any help?

Update: Given the insightful comments below, it is clear it is not converging, hence the actual question now is to find

$$\lim_{x\to 1^-}\left(\sqrt{\frac{\pi}{1-x}}-\sum_{n\geq 1}\frac{x^n}{\sqrt{n}}\right)$$

• And... what did you try? – Did May 17 '16 at 12:56
• Are you sure $\sqrt{\frac{2\pi}{1-x}}$ is not meant to be $\sqrt{\frac{\pi}{1-x}}$? As you have it at the moment the coefficients of $x^n$ differ by a factor of approx $\sqrt2$ for large $n$. – almagest May 17 '16 at 13:09
• I think the stirling formula helps. – user90369 May 17 '16 at 13:11
• Notice $\displaystyle\;\sum_{k=1}^\infty \frac{x^k}{\sqrt{k}} = {\rm Li}_{1/2}(x)$ where ${\rm Li}_s(x)$ is the Polylogarithm function. We have $${\rm Li}_s(e^\mu) = \Gamma(1-s)(-\mu)^{s-1} + \sum_{k=0}^\infty \frac{\zeta(s-k)}{k!} \mu^k \quad\text{ for }\quad |\mu| < 2\pi,\; s \ne 1, 2, 3,\ldots$$ This implies $$\lim_{x\to 1^{-}} \left(\sqrt{\frac{\pi}{1-x}} - \sum_{k=1}^\infty\frac{x^k}{\sqrt{k}}\right) = -\zeta(1/2) \approx 1.460354508809586812889499...$$ – achille hui May 17 '16 at 14:52
• @JackD'Aurizio technically, this doesn't answer the question as it is not a proof. – achille hui May 17 '16 at 14:55

Following the comment posted by @Achillehui, we recognize that the series $\sum_{k=1}^\infty \frac{x^k}{\sqrt{k}}$ is a series representation of the polylogarithm function $\text{Li}_{1/2}(x)$, for $|x|<1$.

An alternative series representation (SEE HERE) of $\text{Li}_{1/2}(x)$ is given by

$$\text{Li}_{1/2}(x)=\sqrt{\frac{\pi}{-\log(x)}}+\sum_{k=0}^\infty \frac{\zeta(1/2-k)}{k!}\log^k(x) \tag 1$$

for $|x|<e^{2\pi}$.

Note that we can expand the logarithm function around $x=1$ as

\begin{align} -\log(x)&=(1-x)\left(1+\sum_{k=2}^\infty\frac{(-1)^{k-1}(x-1)^{k-1}}{k}\right)\\\\ &=(1-x)\left(1+O\left((x-1)\right)\right) \tag 2 \end{align}

Using $(2)$ in $(1)$, we obtain

$$\text{Li}_{1/2}(x)=\sqrt{\frac{\pi}{1-x}}++\zeta(1/2)+O\left(\sqrt{1-x}\right)+\sum_{k=1}^\infty \frac{\zeta(1/2-k)}{k!}\log^k(x) \tag 3$$

Finally, it is easy to see from $(3)$ that

$$\lim_{x\to 1^-}\left(\sqrt{\frac{\pi}{1-x}}-\sum_{k=1}^\infty \frac{x^k}{\sqrt{k}}\right)=-\zeta(1/2)$$

And we are done!

• (+1) How to prove that asymptotic expansion, however, is a tougher problem than the OP's one :) – Jack D'Aurizio May 17 '16 at 16:34
• @JackD'Aurizio Jack, that is for certain. First, we start with an integral representation of $\text{Li}_s(z)$ and deform the Hankel contour. Then, we evaluate that integral as a series of residues. Finally, we fold the series, expand the binomial coefficients, change the order of summation, carry out the inner sum, and arrive at the coveted expression. -Mark – Mark Viola May 17 '16 at 16:49

In order to make user90369's (excellent and elementary) answer a real answer, we have to show that $$\lim_{x\to 1^-}\sqrt{1-x}\sum_{n\geq 1}\frac{x^n}{\sqrt{n}}=\sqrt{\pi}\tag{1}$$ or: $$\lim_{x\to 1^-} (1-x)\sum_{m\geq 2}\left(\sum_{n=1}^{m-1}\frac{1}{\sqrt{n}\sqrt{m-n}}\right)x^m=\pi\tag{2}$$ but that follows from the same argument it is usually exploited to prove Hilbert's inequality, namely that for large $m$s, $$\sum_{n=1}^{m-1}\frac{1}{\sqrt{n}\sqrt{m-n}}=\frac{1}{m}\sum_{n=1}^{m-1}\frac{1}{\sqrt{\frac{n}{m}}\sqrt{1-\frac{n}{m}}}\approx \int_{0}^{1}\frac{dx}{\sqrt{x}\sqrt{1-x}}=\pi.\tag{3}$$ If we set $f(x)=\sum_{n\geq 1}\frac{x^n}{\sqrt{n}}$ and $h(x)=\sqrt{\frac{\pi}{1-x}}$, $(1)$ alone is not enough to ensure that $h(x)-f(x)$ is bounded in a left neighbourhood of $x=1$. However, the study of the Taylor coefficients is:

$$\begin{eqnarray*}h(x)-f(x)&=&\sum_{n\geq 1}\left(\frac{\sqrt{\pi}}{4^n}\binom{2n}{n}-\frac{1}{\sqrt{n}}\right)x^n\\&=&\sum_{n\geq 1}\left(\frac{\Gamma\left(n+\frac{1}{2}\right)}{\Gamma(n+1)}-\frac{1}{\sqrt{n}}\right)x^n\\&\ll&\sum_{n\geq 1}\frac{1}{n\sqrt{n}}\ll 1\tag{1bis}\end{eqnarray*}$$ by Gautschi's inequality. Now we may exploit an analytic continuation. $$f(x)=\sum_{n\geq 1}\frac{x^n}{\sqrt{n}} = \sum_{n\geq 1}\frac{(-1)^{n+1} x^n}{\sqrt{n}}+2\sum_{n\geq 1}\frac{x^{2n}}{\sqrt{2n}}=g(x)+\sqrt{2}\,f(x^2)$$ hence: $$g(x)=\sum_{n\geq 1}\frac{(-1)^{n+1} x^n}{\sqrt{n}} = f(x)-\sqrt{2}\,f(x^2)\tag{4}$$ while $h(x)=\sqrt{\frac{\pi}{1-x}}$ fulfills: $$h(x)-\sqrt{2}\,h(x^2) = h(x)\left(1-\sqrt{\frac{2}{1+x}}\right) \tag{5}$$ hence:

$$\begin{eqnarray*} \lim_{x\to 1^-}\left(h(x)-f(x)\right) &=& \frac{1}{1-\sqrt{2}}\lim_{x\to 1^-}\left[h(x)-\sqrt{2}\,h(x^2)-f(x)+\sqrt{2}\,f(x^2)\right]\\&=&\frac{1}{\sqrt{2}-1}\lim_{x\to 1^-}g(x)\\&=&\frac{1}{\sqrt{2}-1}\sum_{n\geq 1}\frac{(-1)^{n+1}}{\sqrt{n}}=\color{red}{-\zeta\left(\frac{1}{2}\right)}. \tag{6}\end{eqnarray*}$$

• To square, interesting , thank you ! – user90369 May 17 '16 at 15:21
• Leakages from a talented mathematician... – guest May 17 '16 at 15:32
• So your (1) to (3) should be deleted... I dont see any connection between them and the real solution...And is the zeta function continuous at 1/2, I'm not very familiar with zeta functions... – Peter May 17 '16 at 15:48
• @Yongyong: $(1)$ is necessary to make $(3),(4),(5)$ actually work, and to prove $(1)$ I used $(2)$ and $(3)$, so no, they cannot be deleted. And if you are not that confident with the $\zeta$ function, you may simply ignore the very last expression: the answer is given by a (slowly) convergent series. – Jack D'Aurizio May 17 '16 at 15:50
• your result $(1)$ reminds me of the formula $$\lim_{x \to 1^{-1}}\sqrt{1 - x}(x + x^{4} + x^{9} + \cdots) = \int_{0}^{\infty}e^{-x^{2}}\,dx$$ +1 – Paramanand Singh May 18 '16 at 7:36

$f(x):=\sum\limits_{k=1}^\infty\frac{x^k}{\sqrt{k}}$

$g(x):= \sqrt{\frac{2\pi}{1-x}}-f(x)$

$\sum\limits_{k=1}^\infty (-1)^{k-1} \frac{x^k}{\sqrt{k}}=f(x)-\sqrt{2}f(x^2)$ is convergent for $x\uparrow 1$ .

=>

$\sum\limits_{k=1}^\infty (-1)^k \frac{x^k}{\sqrt{k}}+\sqrt{\frac{2\pi}{1-x}}-\sqrt{2}\sqrt{\frac{2\pi}{1-x^2}}= g(x)-\sqrt{2}g(x^2)$.

For $x\uparrow 1$ we get $\sum\limits_{k=1}^\infty (-1)^k \frac{1}{\sqrt{k}}=(1-\sqrt{2})g(1)$.

Therefore is the limit $g(1)=\frac{1}{\sqrt{2}-1}\sum\limits_{k=1}^\infty (-1)^{k-1} \frac{1}{\sqrt{k}}$.

EDIT: See above, it's not $\sqrt{2\pi}$, it's $\sqrt{\pi}$

Thank you all for your efforts!

EDIT 2: $g(1)=-\zeta(\frac{1}{2})$

• When $\sqrt{2\pi}$ really comes into play? – Jack D'Aurizio May 17 '16 at 14:49
• A very good question ! I only assume, that $g(1)$ is $<\infty$, but than it works. Why ? – user90369 May 17 '16 at 14:52
• @Jack D'Aurizio: It seems to be that I don't proof, THAT there is a limit. But IF a limit exits, THEN its $g(1)$. Therefore my question is still opened, you are right. – user90369 May 17 '16 at 15:02
• I just added a proof of the "missing part" in your argument. Now your answer with my prologue gives an elementary alternative to achille hui's technique (asymptotics for $\text{Li}_s$). – Jack D'Aurizio May 17 '16 at 15:18
• I know, thanks again! – user90369 May 17 '16 at 15:41

Using the binomial series you know that $\sqrt{\frac{2\pi}{1-x}}=\sqrt{2\pi}+\sum_{k=1}^\infty\begin{pmatrix}-\frac{1}{2}\\k\end{pmatrix}(-x)^k=\sqrt{2\pi}+\sqrt{2\pi}\sum_{k=1}^\infty \frac{(2k)!}{4^k(k!)^2}x^k$. Using Stirling's formula you know that $\sqrt{2\pi}\frac{(2k)!}{4^kk!}=\frac{\sqrt{2\pi}}{4^k}(1+\epsilon_k)\frac{\sqrt{2\pi(2k)}(\frac{2k}{e})^{2k}}{(\sqrt{2\pi k}(\frac{k}{e})^{k})^2}=(1+\epsilon_k)\frac{\sqrt{2}}{\sqrt{k}}$, where $\epsilon_k\rightarrow 0$ as $k\rightarrow\infty$. Now compare this to $\frac{1}{\sqrt{k}}$, for $k$ large enough, the kompnents of the sum is for instance larger than $\frac{1}{3\sqrt{k}}$, Then the sum is $\geq C+\sum_{k>N}\frac{1}{3\sqrt{k}}x^k$. The right hand side goes to infinity as $x\rightarrow 1$ form left side.

I think the problem would be more interesting if we have $\sqrt{\pi}$ in the first komponent, then it seems that one should modify $\epsilon_k$ to obtain the rate, but this not clear for me, at least this is not readable from the given Stirling's formula.

• Thank you for your efforts! – user90369 May 17 '16 at 14:53