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

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


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.
