Ok, so the instructions state to apply Taylor's Theorum to find the power series (centered at c) for the function, and find the radius of convergence.

Problem: $\displaystyle f(x)=x^{1/2}$ with the center at $\displaystyle c=1$

I took the first through the fourth derivative and then evaluated them at the center to get an idea of the pattern, here's what I got:

1st Derivative: $\displaystyle f'(x)=\frac{1}{2x^{1/2}}$ Evaluation: $\displaystyle f'(1)=\frac{1}{2}$

2nd Derivative: $\displaystyle f''(x)=-\frac{1}{4x^{3/2}}$ Evaluation: $\displaystyle f''(1)=-\frac{1}{4}$

3rd Derivative: $\displaystyle f'''(x)=\frac{3}{8x^{5/2}}$ Evaluation: $\displaystyle f'''(1)=\frac{3}{8}$

4th Derivative: $\displaystyle f''''(x)=-\frac{15}{16x^{7/2}}$ Evaluation: $\displaystyle f''''(1)=-\frac{15}{16}$

I can use the Taylor series formula easily enough, however, I'm not sure how they came up with the answer for $\displaystyle f^n(1)=(-1)^{n-1}\frac{1*3*....*5*(2n-3)}{2^n}$ which holds true when n is equal to or greater than 2.

Now, I can write up the power series and perform the ratio test to find the radius of convergence on my own easily enough, however, I'm unfamiliar with this notation. I understand the $\displaystyle (-1)^{n-1}$ in the answer because that must be true for the signs to alternate back and forth. Due to the denominators in each evaluation, I can also see where $\displaystyle 2^n$ comes into play since the denominator is just 2 raised to an exponent equal to n. What I don't see is where that frickin' $\displaystyle 1*3*....*5*(2n-3)$ comes into play. Can anyone explain to me what that notation is, why it's part of the answer and how I might have to treat it when taking a limit? Thanks in advance guys.