Calculus Infinite Series Homework Problem. I have the problem
"Let
$f(x) = x + \frac{2}{3} x^3 + \frac{2 \cdot 4}{3 \cdot 5} x^5 + \dots + \frac{2 \cdot 4 \dotsm 2n}{3 \cdot 5 \dotsm (2n + 1)} x^{2n + 1} + \dotsb$
on the interval $(-1,1)$ of convergence of the defining series.
(a) Prove that $(1 - x^2) f'(x) = 1 + xf(x).$
(b) Prove that $f(x) = \frac{\arcsin x}{\sqrt{1 - x^2}}.$"
I don't exactly know how to prove it without doing part b first, then part a. Would prefer if someone could prove it using the infinite series. If anyone can help me with proving part a, then part b, that'd be greatly appreciated.
 A: If you are given a power series $f(x) = \sum_{n = 0}^\infty a_n (x - x_0)^n$ with radius of convergence $R$, then $f$ is differentiable on $(x_0 - R, x_0 + R)$ and the derivative is $f'(x) = \sum_{n = 1}^\infty n a_n (x - x_0)^{n - 1}$.
In other words, you can differentiate term-by-term.
I shall leave it to you to verify that your series does converge on $(-1, 1)$. (For example, use an appropriate version of the root test.)

Recall the notion of a double factorial: We have $$(2n)!! = 2 \cdot 4 \cdot \cdots \cdot (2n)$$ and $$(2n + 1)!! = 1 \cdot 3 \cdot \cdots \cdot (2n + 1).$$
By convention, we have $(-1)!! = 0!! = 1$.
In particular, for $n \ge 1$, we have the identity $\frac{n!!}{n} = (n - 2)!!$.

Now, in your case, we have
$$f(x) = \sum_{n = 0}^\infty \frac{(2n)!!}{(2n + 1)!!}x^{2n + 1}.$$
and thus,
\begin{align} 
f'(x) &= \sum_{n = 0}^\infty \frac{(2n)!!}{(2n + 1)!!} \cdot (2n + 1) \cdot x^{2n} \\ 
&= 1 + \sum_{n = 1}^\infty \frac{(2n)!!}{(2n - 1)!!} x^{2n}.
\end{align}
Thus, we get
\begin{align} 
x^2f'(x) &= x^2 + \sum_{n = 1}^\infty \frac{(2n)!!}{(2n - 1)!!} x^{2n + 2} \\
&= x^2 + \sum_{n = 2}^{\infty} \frac{(2n - 2)!!}{(2n - 3)!!} x^{2n} \\
&=\sum_{n = 1}^{\infty} \frac{(2n - 2)!!}{(2n - 3)!!} x^{2n}.
\end{align}
Subtracting from the earlier expression, we get
\begin{align} 
(1 - x^2)f'(x) &= 1 + \sum_{n = 1}^{\infty}\left[\frac{(2n)!!}{(2n - 1)!!} - \frac{(2n - 2)!!}{(2n - 3)!!}\right] x^{2n}.
\end{align}
On the other hand, we have the expression
\begin{align} 
1 + xf(x) &= 1 + \sum_{n = 0}^\infty \frac{(2n)!!}{(2n + 1)!!}x^{2n + 2} \\
&= 1 + \sum_{n = 1}^{\infty} \frac{(2n - 2)!!}{(2n - 1)!!} x^{2n}.
\end{align}
I leave it to you to check that
$$\frac{(2n)!!}{(2n - 1)!!} - \frac{(2n - 2)!!}{(2n - 3)!!} =  \frac{(2n - 2)!!}{(2n - 1)!!}$$
for all $n \ge 1$. Thus, we have shown that
$$(1 - x^2)f'(x) = 1 + xf(x).$$

Now, we wish to conclude that $f(x) = \frac{\arcsin x}{\sqrt{1 - x^2}}$. For this, simply note that $g(x) := \frac{\arcsin x}{\sqrt{1 - x^2}}$ also satisfies the first order ODE
$$y' - \frac{x}{1 - x^2}y = \frac{1}{1 - x^2}$$
on $(-1, 1)$.
The above is a linear first-order ODE with continuous coefficients on $(-1, 1)$. Thus, it has a unique solution which satisfies $y(0) = 0$.
Since $f$ and $g$ both satisfy the ODE with same initial condition, we see that $f = g$. Thus,
$$\sum_{n = 0}^\infty \frac{(2n)!!}{(2n + 1)!!}x^{2n + 1} = \frac{\arcsin x}{\sqrt{1 - x^2}}.$$
