Taylor Series Expansion of $ f(x) = \sqrt{x} $ around $ a = 4 $ guys.
Here's the exercise: find a series representation for the function $ f(x) = \sqrt{x} $ around $ a = 4 $ and find it's radius of convergence.
My doubt is on the first part: I can't seem to find a pattern.
$ \circ f(x) = \sqrt x \rightarrow f(4) = 2 \\ \circ f'(x) = \frac{1}{2} \cdot \frac{1}{\sqrt x} \rightarrow f'(4) = \frac{1}{4} \\ \circ f''(x) = \frac{1}{2} \cdot \left( -\frac{1}{2} \right) \cdot x^{-\frac{3}{2}} \rightarrow f''(4) = \frac{-1}{4 \cdot 8} = \frac{-1}{4 \cdot (2 \cdot 4^1)} = \frac{-1}{2 \cdot 4^2} \\\\ \circ f^{(3)}(x) = -\frac{1}{4} \cdot \left(- \frac{3}{2} \right) \cdot x^{-\frac{5}{2}} \rightarrow f^{(3)}(4) = \frac{3}{4 \cdot 32} = \frac{3}{4 \cdot (4^2 \cdot 2)} = \frac{3}{4^3 \cdot 2} \\\\ \circ f^{(4)}(x) = \frac{3}{8} \cdot \left(- \frac{5}{2} \right) \cdot x^{-\frac{7}{2}} \rightarrow f^{(4)}(4) = \frac{-3 \cdot 5}{4 \cdot 128} = -\frac{3 \cdot 5}{4 \cdot (4^3 \cdot 2)} = -\frac{3 \cdot 5}{4^4 \cdot 2} $
hmm.. Can I say then that:
$ f(x) = 2 + \frac{(x-4)}{4} + \frac{1}{2} \cdot \sum_{n=2}^{\infty} \frac{(-1)^{n-1} \cdot (1 \cdot 3 \cdot 5 \dots (2n-3)) \cdot (x-4)^n}{4^n} $ and using the ratio test:
$ \lim_{n \to \infty} \left| \frac{(x-4)^{n+1} \cdot (2n-1)}{4^{n+1}} \cdot \frac{4^n}{(x-4)^n \cdot (2n-3)} \right| = \frac{|x-4|}{4} \rightarrow -4 < x-4 < 4 \therefore 0 < x < 8 $ and thus R = 4.
But here's the thing: the answer on the book is:
$ 2 + \frac{(x-4)}{4} + 2 \cdot \sum_{n=2}^{\infty} (-1)^{n-1} \cdot \frac{1 \cdot 3 \cdot 5 \dots (2n-3)  \cdot (x-4)^n}{2 \cdot 4 \cdot 6 \dots (2n) \cdot 4^n} $
Am I missing something? 
Thanks in advance, guys!

You were on the right track.  We have for $n\ge 2$
$$\begin{align}
f^{(n)}(x)&=(-1)^{n-1}\frac12 \frac12 \frac32 \frac52 \cdots \frac{2n-3}{2}x^{-(2n-1)/2}\\\\
&=(-1)^{n-1}\frac{(2n-3)!!}{2^n}x^{-(2n-1)/2}\tag 1
\end{align}$$
where $(2n-3)!! = 1\cdot 3\cdot 5\cdot (2n-3)$ is the double factorial of $2n-3$.  Evaluating $(1)$ at $x=4$ yields
$$f^{(n)}(4)=(-1)^{n-1}\frac{2(2n-3)!!}{2^n4^n}$$
Therefore, we have
$$\begin{align}
f(x)&=f(4)=f'(4)(x-4)+2\sum_{n=2}^{\infty}(-1)^{n-1}\frac{(2n-3)!!}{(2^n)(4^n)n!}(x-4)^n\\\\
&=2+\frac{(x-4)}{4}+2\sum_{n=2}^{\infty}(-1)^{n-1}\frac{(2n-3)!!}{(2n)!!(4^n)}(x-4)^n\\\\
\end{align}$$
where $(2^n)n!=(2n)!!=2\cdot 4\cdot 6\cdots 2n$ and again $(2n-3)!!=1\cdot3\cdot 5\cdots(2n-3)$.  And we are done!
