Taylor series expansion for $f(x)=\sqrt{x}$ for $a=1$ I seem to be stuck defining an alternating sequence of terms in this series because $f^{(0)}(x)=f(x)$ is positive, as well as $f'(x)$, but then every other term starting with $f''(x)$ is negative. How can I define $f^{(n)}(x)$ given this?
\begin{array}{ll}
    f(x)=x^{\frac{1}{2}} & f(1)=1 \\
    f'(x)=\frac{1}{2}\cdot x^{-\frac{1}{2}} & f'(1)=\frac{1}{2} \\
    f''(x)=(-1)^1\cdot\left(\frac{1}{2}\right)^{2}\cdot x^{-\frac{3}{2}} & f''(1)=(-1)^1\cdot\left(\frac{1}{2}\right)^{2} \\
    f'''(x)=(-1)^2\cdot 3\cdot\left(\frac{1}{2}\right)^{3}\cdot x^{-\frac{5}{2}} & f'''(1)=(-1)^2\cdot3\cdot\left(\frac{1}{2}\right)^{3} \\
    f^{(4)}(x)=(-1)^3\cdot3\cdot 5\cdot\left(\frac{1}{2}\right)^{4}\cdot x^{-\frac{7}{2}} & f^{(4)}(1)=(-1)^3\cdot3\cdot 5\cdot\left(\frac{1}{2}\right)^{4} \\
    f^{(n)}(x)=(-1)^{n-1}\left(\frac{1}{2}\right)^{n}\cdot x^{\frac{1-2n}{2}} & f^{(n)}(1)=(-1)^{n-1}\left(\frac{1}{2}\right)^{n}
\end{array}
I thought I had the right answer until I realized that I'd be defining $f(x)$ to be negative.

Well you're not defining $f(x)$ to be negative, you're finding that certain derivatives of $f(x)$ are negative - if you think about the graph of this $f(x)$ it decreasingly increases; that is, it is increasing, but the rate at which it increases is decreasing, so it should seem to me that in the least, the second derivative should be negative, which is what your formula suggests!
Was in comments then realized this was a long comment :P
EDIT: So thanks to my oversight I wasn't too helpful. Hopefully this edit will provide more help
So if $f(x) = x^{1/2}$ then...
$$
f'(x) = \frac{1}{2} x^{-1/2} \\
f''(x) = \frac{-1}{4} x^{-3/2} \\
f'''(x) = \frac{3}{8} x^{-5/2} \\
f^{(4)} (x) = \frac{-15}{16} x^{-7/2} \\
f^{(5)} (x) = \frac{105}{32} x^{-9/2}
$$
so, at least to me, it appears as though you get $f^{(0)} (x) = x^{1/2}$, $f^{(1)} (x) = \frac{1}{2} x^{-1/2}$ and $f^{(n)} (x) = (-1)^{n+1} \frac{(2n - 3)(2n - 5) \cdots (3)(1)}{2^n} x^{\frac{1-2n}{2}} : n \ge 2$. At least for me (I can be wrong though) I don't see any outstanding pattern other than the one I just said. Hope this helps!
