$$\frac{x-2}{\sqrt{x^2-4x+8}} \quad \text{at}\ x_0=2$$
Let $t = x-2$. then as $x\to 2$, $t\to 0$ and we can find the Maclaurin Series in terms of $t$.
\begin{align*} \frac{x-2}{\sqrt{x^2-4x+8}} &=\frac{x-2}{\sqrt{(x-2)^2+4}}\\ &=\frac{t}{\sqrt{t^2+4}}\\ &=\frac{1}{2}\frac{t}{\sqrt{1+\left(\frac{t}{2}\right)^2}} \end{align*} Since the Maclaurin Series
$$ \left(1+\left(\frac{t}{2}\right)^2\right)^{-1/2} = \sum_{k=0}^n{-1/2 \choose k}\left(\frac{t}{2}\right)^{2k} + o(t^{2n})$$
Then
$$\frac{t}{2}\left(1+\left(\frac{t}{2}\right)^2\right)^{-1/2} = \sum_{k=0}^n{-1/2 \choose k}\left(\frac{t}{2}\right)^{2k+1} + o(t^{2n})$$
Here we have
\begin{align*} {-1/2 \choose k} &= \frac{-\frac{1}{2}\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right)...\left(-\frac{1}{2}-(k-1)\right)}{k!}\\ &=\frac{-\frac{1}{2}\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)...\left(-\frac{2k-1}{2}\right)}{k!} \end{align*}
If we denote $1\cdot 3\cdot 5\cdot ... \cdot (2k-1) = (2k-1)!!$ we get
$${-1/2 \choose k} = \frac{(-1)^k(2k-1)!!}{2^k k!}$$
And finally
\begin{align*}\frac{t}{2}\left(1+\left(\frac{t}{2}\right)^2\right)^{-1/2} &= \sum_{k=0}^n{-1/2 \choose k}\left(\frac{t}{2}\right)^{2k+1} + o(t^{2n})\\ &= \frac{t}{2} + \sum_{k=1}^n\frac{(-1)^k(2k-1)!!}{2^kk!}\left(\frac{t}{2}\right)^{2k+1} + o(t^{2n}) \end{align*}
And in terms of $x$:
$$ \frac{x-2}{\sqrt{x^2-4x+8}}\Bigg\rvert_{x_0=2} = \frac{x-2}{2} + \sum_{k=1}^n\frac{(-1)^k(2k-1)!!}{2^kk!}\left(\frac{x-2}{2}\right)^{2k+1} + o((x-2)^{2n})$$
But the textbook's answer is $$ \frac{x-2}{2} + \sum_{k=1}^{n-1}\frac{(-1)^k(2k-1)!!}{2^{2k+1}(2k)!!}(x-2)^{2k+1} + o((x-2)^{2n})$$
I don't get why the sum goes just to $n-1$ instead of $n$ and instead of $2^{3k+1}k!$ in the denominator, as I got, it is $2^{2k+1}(2k)!!$
What am I missing?
There's no error in what you did, you only "wasted" two orders of magnitude in your remainder term.
The terms of the textbook's answer look different from yours, but are in fact the same, since
$$(2k)!! = 2^k\cdot k!$$
for one definition of the double factorial of even natural numbers (there's another definition that differs from the used one by a factor of $\sqrt{\frac{2}{\pi}}$), thus
$$2^{3k+1}\cdot k! = 2^{2k+1}(2k)!!$$
and consequently we see that the terms are identical.
You wasted one order of magnitude in the remainder term by writing
$$\biggl(1 + \Bigl(\frac{t}{2}\Bigr)^2\biggr)^{-1/2} = \sum_{k = 0}^n \binom{-\frac{1}{2}}{k} \biggl(\frac{t}{2}\biggr)^{2k} + o(t^{2n})\tag{$\ast$}$$
and interpreting the polynomial as the Taylor polynomial of order $2n$. It is also the Taylor polynomial of order $2n+1$, and therefore one can write the remainder term as $o(t^{2n+1})$. The next order of magnitude you wasted in the remainder term was when you multiplied $(\ast)$ with $\frac{t}{2}$, since $\frac{t}{2}o(t^m) = o(t^{m+1})$, and you didn't increase the exponent.
So with these orders taken into account, we find
$$\frac{x-2}{\sqrt{x^2-4x+8}} = \frac{x-2}{2} + \sum_{k = 1}^m \frac{(-1)^k (2k-1)!!}{2^{2k+1}(2k)!!}(x-2)^{2k+1} + o\bigl((x-2)^{2m+2}\bigr)$$
for any fixed $m\in \mathbb{N}$. So if we want to have a remainder term in $o(t^{2n})$, it suffices to take $m = n-1$.