finding the taylor series for a function so im trying to find the taylor series representation of $\frac{1}{\sqrt{x}}$ around $1$.
i managed to find this so far:
$$
\sum_{n=0}^\infty \frac{(x-1)^n \cdot (-1)^n }{  2^n \cdot n!}
$$
the last thing im missing is in the numerator- but i cant figure out what exactly and how to represent it. Any ideas?
 A: By using the binomial theorem, one has, for $|t|<1$,
$$
(t+1)^{-1/2}=1-\tfrac12t+\tfrac12\tfrac32\cdot t^2-\tfrac12\tfrac32\tfrac52\cdot t^3+\dots+(-1)^n\tfrac{1\cdot2\cdot3\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}\cdot t^n+\cdots
$$ or, by putting $t=x-1$, as $x \to 1$
$$
\begin{align}
\frac1{\sqrt{x}}&=\frac1{\sqrt{t+1}}
\\\\&=1+\sum_{n=1}^\infty\:(-1)^n\cdot\frac{1\cdot2\cdot3\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}\cdot(x-1)^{n}
\\\\&=1+\sum_{n=1}^\infty\:(-1)^n\cdot\frac{(2n)!}{2^{2n} (n!)^2}\cdot(x-1)^{n}.
\end{align}
$$
A: Why not just apply the binomial theorem? We note that $$\frac{1}{\sqrt{x}} = \frac{1}{\sqrt{1+(x-1)}} = [1+(x-1)]^{-1/2} =\sum_{n=0}^\infty(x-1)^n{-1/2 \choose n} $$
Note that we can turn this into your form by using the definition of Binomial Coefficients
A: If you're just looking for some compact notation, let me introduce to you the double factorial:  for $n$ odd, 
$$
n!! = n \cdot (n-2) \cdot (n - 4) \cdots 1.
$$
So, if you take $a_n = (2n - 1)!!$, you have $a_1 = 1, a_2 = 3, a_3 = 15, a_4 = 105, a_5 = 1575$, and so forth;  this is exactly the sequence of odd coefficients that appears in the power series coefficients.
