Series expansion of $\sqrt{x}$ I am interested in a series expansion of $\sqrt{x}$ in powers of $x$ for real $x$ around some arbitrary point $0<a<1$ (with the respective radius of convergence for $x$). I start with the well known Taylor series for
$$ \sqrt{1+x'} = \sum_{n=0}^{\infty} \binom{\frac{1}{2}}{n}x'^n,$$ for $x'\in(-1,1)$. Then I do the transformation $x'\rightarrow x - a $, for $a\in(0,1)$ to arrive at a double sum
$$ \sum_{0\le i \le n}^{\infty} \binom{\frac{1}{2}}{n}\binom{n}{i}x^i(-a)^{n-i}.$$
Now I like to rearrange the thing for powers of $x$, after some gymnastics I arrive at
$$\sum_{m=0}^{\infty} x^m \sum_{0\le j \le m} \binom{\frac{1}{2}}{m+j}\binom{m+j}{j}(-a)^j.$$ At this point I feel that the second sum could be simplified further, but I could not succeed in doing so, yet. 
Can this expression be further simplified? 
$\mathbf{Note:}$ An interesting special case would be the algebraic expression for the limit $a\rightarrow 1^-$, as it would represent a series expansion of $\sqrt{x}$ around $x=0$ (which is not possible because $\sqrt{x}$ is not differentiable at 0). 
 A: For the case of $a=1$:
Just as Tom Cooney, it is not very clear to me. 
Anyway, using a CAS, what I got is $$a_m=\sum_{j=0}^m(-1)^j \binom{\frac{1}{2}}{m+j}\binom{m+j}{j}=-\frac{2 (-1)^m (m+1) \binom{\frac{1}{2}}{2 m+1} \binom{2 m+1}{m+1}}{2 m-1}$$ The problem is that $$\Phi(x)=\sum_{m=0}^\infty a_mx^m=\, _3F_2\left(-\frac{1}{2},\frac{1}{4},\frac{3}{4};\frac{1}{2},2;-4
   x\right)+\frac{3}{8} x \,
   _3F_2\left(\frac{1}{2},\frac{5}{4},\frac{7}{4};\frac{3}{2},3;-4 x\right)$$ where appear  the generalized hypergeometric function.
Edit
There is something interesting to notice : noting $f_k$ the value of the last given expression for $x_k=10^{2k}-1$, the following values are obtained $$f_1=10.1250000$$ $$f_2=100.0506048$$ $$f_3=1000.017094$$ $$f_4=10000.00551$$ $$f_5=100000.0018$$ which are closer and closer to $\sqrt{x_k+1}$.
For infinite values of $x$, the asymptotics of $\Phi(x)$ is given by $$\Phi(x)=\sqrt{x}-\frac{3 \,\Gamma \left(-\frac{3}{4}\right)}{16\,\sqrt{\pi
   }\ \Gamma \left(\frac{7}{4}\right)}\frac{1}{\sqrt[4]{x}}+O\left(\frac{1}{x^{3/4}}\right)$$ If, as discussed on chat, you really want to see $\sqrt{x+1}$, the asymptotic would be $$\Phi(x)=\sqrt{x+1}+\frac{\Gamma \left(\frac{5}{4}\right)}{\sqrt{\pi } \Gamma \left(\frac{7}{4}\right)}\frac{1}{\sqrt[4]{x}}+O\left(\frac{1}{x^{1/2}}\right)$$
