$Q^1$: Given that
\begin{align} a_0 &= 0\\ a_{n+1} &= na_n+1\\ \end{align}
find the closed form$?$
My attempt:
\begin{align}
\sum_{n\geq0}^{\infty}a_{n+1}x^{n+1}&=\sum_{n\geq0}^{\infty}na_nx^{n+1}+\sum_{n\geq0}^{\infty}1.x^{n+1}\\
(G(z)-a_0)&=G'(z)x^2+\frac{x}{1-x} \qquad\text{Using }G'(z)x=\sum_{n\geq0}^{\infty}na_nx^n
\end{align}
But how to deal with $G'(z)?$
$Q^2$: Let $\{C_n\}$ be the sequence of Catalan numbers, that is, the solution to the recurrence relation $C_n=\sum_{k=0}^{n-1}C_kC_{n-k-1}$ with $C_0=C_1=1$
$(a)$ Show that if $G(x)$ is the generating function for the sequence of Catalan numbers, then $xG(x)^2-G(x)+1=0$ concluding $G(x)=\frac{1-\sqrt{1-4x}}{2x}$.
$(b)$ Conclude that $G(x)=\sum_{n=0}^{\infty}\frac{1}{n+1}\binom{2n}{n}x^n$ so that $C_n=\frac{1}{n+1}\binom{2n}{n}$
My attempt:
I solve $(a)$ but stuck in $(b)$. The solution provided said that
Using the extended binomial theorem: $$(1-4x)^{-\frac{1}{2}}=\sum_{k=0}^{\infty}\binom{2k}{k}x^k$$ $\color{red}{\text{Note that}(1-4x)^{-\frac{1}{2}}\text{is the derivative of}G(x)=\frac{1-\sqrt{1-4x}}{2x},\text{thus }xG(x)\text{is the integral of}(1-4x)^{-\frac{1}{2}}}$ $$xG(x)=\int_0^x(1-4x)^{-\frac{1}{2}}dx=\cdots=\sum_{n=0}^{\infty}\frac{1}{n+1}\binom{2n}{n}x^n$$ Hence, $C_n=\frac{1}{n+1}\binom{2n}{n}$.
I didn't understood the red line. I know asking one question at a time but since these two question related with derivative and integration hence putted on a single thread.
Thanks in advance and thanks for your time .
$Q^1$: I've played around a bit with initial conditions and indexing and arrived at the following solution. This is just the OEIS sequence A000522 with a $0$ tacked on in the front.
We will use exponential generating functions. First let $b_n = a_{n+1}$. Then we have $$b_{n} = nb_{n-1} + 1; \qquad b_0 = 1$$
Let $B(x) = \sum_{n\ge 0} b_n \frac{x^n}{n!}$, then
$$\sum_{n\ge 0} b_{n}\frac{x^n}{n!} = \sum_{n \ge 0 }nb_{n-1} \frac{x^n}{n!} + \sum_{n \ge 0} \frac{x^n}{n!}$$
which yields \begin{align} B(x) &= x\sum_{n \ge 0 }b_{n-1} \frac{x^{n-1}}{(n-1)!} + e^x = xB(x) + e^x \end{align} and so we obtain the g.f. $B(x) = \frac{e^x}{1-x}$. Using the series expansions for $e^x$ and $(1-x)^{-1}$ we have
\begin{align} \frac{e^x}{1-x} &= \left(\sum_{n\ge 0} \frac{x^n}{n!}\right)\left(\sum_{n\ge 0} x^n\right)\\ &= \sum_{n\ge 0} \left(\sum_{j=0}^{n} \frac{1}{j!} \cdot 1\right) x^n\\ &= \sum_{n\ge 0} \left(\sum_{j=0}^{n} \frac{n!}{j!} \right) \frac{x^n}{n!}. \end{align} (The product of $2$ polynomials $\left(\sum_{n\ge 0} s_n x^n\right)\left(\sum_{n\ge 0} t_n x^n\right) = \sum_{n\ge 0} \left(\sum_{j=0}^{n} s_jt_{n-j}\right)x^n$)
So our closed formula is $b_n = \sum_{j=0}^n \frac{n!}{j!}$, and to go back to our original problem we have $$a_n = \begin{cases} 0 &n=0\\ \sum_{j=0}^n \frac{n!}{j!} &n\ge 1 \end{cases}$$
Well, it depends on what you really mean by closed formula. If someone has a less hacky solution I'm all ears.
$Q^2$: For other answers to this, you may want to see this the stack exchange question (specifically rogerl's answer for a more algebraic solution).
I believe the solution (or you) may have a typo. Note that $(1-4x)^{-\frac{1}{2}}$ is the derivative of $xG(x) = \frac{1-(1-4x)^{\frac{1}{2}}}{2}$, or in other words by the Fundamental Theorem of Calculus: $$xG(x) = \frac{1-(1-4x)^{\frac{1}{2}}}{2} = \int_0^x(1-4t)^{-\frac{1}{2}} dt$$ which then allows us to integrate $\sum_{k=0}^\infty \binom{2k}{k}x^k$ term by term to obtain $$xG(x) = \sum_{k=0}^\infty \binom{2k}{k}\frac{1}{k+1}x^{k+1}$$ and dividing both sides by $x$ gives us our desired solution.