I would really appreciate some help on this problem which I've been stuck on. I can't seem to derive the interval of convergence of the Taylor series for square root $x$.
I know the Taylor series can be expressed as such:
\begin{align} p(x) &= \sum_{n=0}^\infty \dfrac{f^{(n)}(4)}{n!}(x - 4)^n\\ &= \sum_{n=0}^\infty \dfrac{(-1)^{n-1}[(2n - 3)!!](4)^\frac{1}{2}}{(2(4))^n}\dfrac{1}{n!}(x - 4)^n\\ &= \sum_{n=0}^\infty \dfrac{(-1)^{n-1}[(2n - 3)!!](4)^\frac{1}{2}}{(8)^n}\dfrac{1}{n!}(x - 4)^n \end{align}
And I tried using the ratio test for convergence:
\begin{align} \dfrac{|u_{n+1}|}{|u_n|} &= \dfrac{\dfrac{|(-1)^{n}(2n - 1)!!(2)(x - 4)^{n+1}|}{|(8)^{n+1}\cdot(n+1)!|}}{\dfrac{|(-1)^{n-1}(2n - 3)!!(2)(x - 4)^{n}|}{|(8)^{n}\cdot(n)!|}}\\ \text{which I simplified to:}\\ &= \dfrac{(2n - 2)!!}{8(n + 1)(2n - 3)!!}(x - 4)\\~\\ \lim_{n\to\infty} \dfrac{|u_{n+1}|}{|u_n|} &= \lim_{n\to\infty}\dfrac{(2n - 2)!!}{8(n + 1)(2n - 3)!!}(x - 4) \end{align}
but I couldn't evaluate the limit :(
I don't know why I can't seem to solve this, but I know there must be a way since Wolfram Alpha says the series converges for $\mid(1-x/4)\mid<1 $