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

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


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