find the interval of convergence of Taylor series Represent the function $f(x)= x^{0.5}$ as a power series: $\displaystyle \sum_{n=0}^\infty c_n(x−6)^n$
Got that: $c_0$ = $\sqrt{6}$
$C_1=\dfrac{1}{2\sqrt{6}}$
...
But I couldn't find the interval of convergence. I thought we'd require $|x-6| < 1$.
 A: Note that
$$
\sqrt{6+h}=f(6+h)=\sum_{n=0}^\infty \frac{f^{(n)}(6)}{n!}(x-6)^n,\tag{1}
$$
where, for $n\ge 1$,
$$
f^{(n)}(a)=\frac{(-1)^{n-1}a^{-(2n-1)/2}\prod_{k=1}^n (2k-1)}{2^n}=
\frac{(-1)^{n-1}a^{-(2n-1)/2}(2n)!}{4^n n!}.
$$
Thus
$$\sqrt{6+h}=\sum_{n=0}^\infty \frac{(-1)^{n-1}6^{-(2n-1)/2}(2n)!}{4^n (n!)^2}(x-6)^n=\sum_{n=0}^\infty a_n(x-6)^n.
$$
Now, using the ratio test
$$
\left|\frac{a_n}{a_{n+1}}\right|=\left|\frac{6\cdot 4 (n+1)^2}{2n(2n-1)}\right|\to 6,
$$
as $n\to\infty$.
Hence the radius of convergence of $(1)$ is $R=6$.
A: Let $f(x) = \sqrt{x}$, and $a = 6$. Apply Taylor expansion formula for $f$ at $x = a$ we have:
$f(x) = f(a) + f'(a)(x-a) + \dfrac{f''(a)}{2!}(x-a)^2 + ....+ \dfrac{f^{(n)}(a)}{n!}(x-a)^n + ... = \sqrt{6} + \dfrac{1}{2\sqrt{6}}(x-6) - \dfrac{1}{48\sqrt{6}}(x-6)^2 + .......+\dfrac{(-1)^{n+1}\cdot \displaystyle \prod_{k=1}^{n-1}(2k-1)\cdot 6^{-(2n-1)/2}}{n!\cdot 2^n}(x-6)^n +......$.
$R = \displaystyle \lim_{n \to \infty} \left|\dfrac{a_{n+1}}{a_n}\right| = \displaystyle \lim_{n \to \infty} \dfrac{2n-1}{12n}\cdot |x-6| = \dfrac{|x-6|}{6} < 1 \iff |x-6| < 6 \iff -6 < x - 6 < 6 \iff 0 < x < 12$.
