<s>[INST] 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 $|x-6| < 1$.
 [/INST] Let $f(x) = \sqrt{x}$, and $a = 6$. Apply Taylor expansion formula for $f$ at $x = a$ we have:
$f(x) = f(a) + f'(a)(x-a) + \dfrac{f''(a)}{2!}(x-a)^2 + ....+ \dfrac{f^{(n)}(a)}{n!}(x-a)^n + ... = \sqrt{6} + \dfrac{1}{2\sqrt{6}}(x-6) - \dfrac{1}{48\sqrt{6}}(x-6)^2 + .......+\dfrac{(-1)^{n+1}\cdot \displaystyle \prod_{k=1}^{n-1}(2k-1)\cdot 6^{-(2n-1)/2}}{n!\cdot 2^n}(x-6)^n +......$.
$R = \displaystyle \lim_{n \to \infty} \left|\dfrac{a_{n+1}}{a_n}\right| = \displaystyle \lim_{n \to \infty} \dfrac{2n-1}{12n}\cdot |x-6| = \dfrac{|x-6|}{6} < 1 \iff |x-6| < 6 \iff -6 < x - 6 < 6 \iff 0 < x < 12$.
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