Consider the function $f(x)=\arcsin{(x)}$ and the task of finding the Taylor series at $0$ for this function.
We have $$f'(x)=\arcsin'{(x)}=\frac{1}{\sqrt{1-x^2}}$$
Let $g(x)=(1+x)^\alpha$. Then, for $|x|<1$ and any $\alpha$,
$$g(x)=\sum\limits_{k=0}^\infty \binom{\alpha}{k} x^k$$
That is, $g$ can be represented as an infinite series, the binomial series.
Note that this infinite series is a convergent power series centered at $0$. Hence, it must be the Taylor series of $g$ at $0$.
Using this fact we can show that for $|x|<1$,
$$\frac{1}{\sqrt{1-x^2}}=\sum\limits_{k=0}^\infty \binom{-1/2}{k}x^{2k}$$
Thus, for $|x|<1$,
$$f'(x)=\arcsin'{(x)}=\frac{1}{\sqrt{1-x^2}}=\sum\limits_{k=0}^\infty \binom{-1/2}{k}x^{2k}$$
If we integrate we obtain the result
$$f(x)=\arcsin{(x)}=\sum\limits_{k=0}^\infty \binom{-1/2}{k}\frac{x^{2k+1}}{2k+1}$$
Which, if true, means we have another convergent power series centered at $0$ so we have found the desired Taylor series of $\arcsin$ at $0$.
My question is: how do we justify this integration step?
If we prove that the sequence of functions
$$\{f_n'\}=\left\{\sum\limits_{k=0}^n \binom{-1/2}{k}x^{2k}\right\}$$
converges uniformly to $\arcsin'$ on some interval, then I know of a theorem that says that
$$\lim\limits_{n\to\infty}\int_a^b f_n' =\int_a^b \arcsin'$$
This seems like it might involve a lot of calculations. Nonetheless, would it be a valid justification?
We could also just use the fact that $\arcsin'$ is continuous and use the first fundamental theorem of calculus.
Is this a correct justification?
For each $n\in\Bbb N$,$$\left|\binom{-\frac12}n\right|=\left|(-1)^n\frac{(2n-1)!!}{(2n)!!}\right|<1,$$and therefore the radius of convergence of the series $\sum_{n=0}^\infty\binom{-1/2}nx^n$ is at least $1$ (it turns out that it is equal to $1$). So, the series converge uniformly on any interval $[-a,a]$, when $a\in(0,1)$, and this uniform convergence justifies the integration step.