<s>[INST] How can I get f(x) from its Maclaurin series? I know how to get a Maclaurin series when $f(x)$ is given. I have to find $\sum_{n=0}^{\infty}\frac{f^{(k)}(0)}{k!}x^k$.
But how can I get $f(x)$ from its Taylor series?
The problem is $$f(x) = \sum_{n=0}^{\infty} C_n x^n,$$ where $C_n$ is a Catalan number defined by $C_n = \frac{1}{n+1}\binom{2n}{n}$.
How can I get $f(x)$?
 [/INST] *

*The Catalan numbers $$C_n=\frac{1}{n+1}\binom{2n}{n}$$ are strongly related with the Central binomial coefficients $\binom{2n}{n}$.

*The binomial identity 
\begin{align*}
\frac{1}{2n-1}\binom{2n}{n}=(-1)^{n+1}\binom{\frac{1}{2}}{n}4^n\tag{1}
\end{align*}
provides a relationship with the binomial series expansion
\begin{align*}
\sqrt{1+x}=\sum_{n= 0}^\infty\binom{\frac{1}{2}}{n}x^n\tag{2}
\end{align*}

From (1) we obtain a relationship with the Catalan numbers.
  \begin{align*}
\frac{1}{2}\cdot\frac{1}{2n-1}\binom{2n}{n}&=\frac{1}{2}\cdot\frac{1}{2n-1}\cdot\frac{2n}{n}\binom{2n-1}{n-1}\\
&=\frac{1}{2n-1}\binom{2n-1}{n-1}\\
&=\frac{1}{2n-1}\binom{2n-1}{n}\\
&=\frac{1}{n}\binom{2n-2}{n-1}\\
&=C_{n-1}\tag{3}
\end{align*}
We conclude
  \begin{align*}
\sum_{n=0}^\infty C_nx^n&=\sum_{n=1}^\infty C_{n-1}x^{n-1}\tag{4}\\
&=\frac{1}{2}\sum_{n=1}^\infty\frac{1}{2n-1}\binom{2n}{n}x^{n-1}\tag{5}\\
&=\frac{1}{2}\sum_{n=1}^\infty(-1)^{n+1}\binom{\frac{1}{2}}{n}4^nx^{n-1}\tag{6}\\
&=-\frac{1}{2x}\sum_{n=1}^\infty\binom{\frac{1}{2}}{n}(-4x)^{n}\tag{7}\\
&=-\frac{1}{2x}\left(\sqrt{1-4x}-1\right)\tag{8}\\
&=\frac{1}{2x}\left(1-\sqrt{1-4x}\right)
\end{align*}

Comment:


*

*In (3) we use twice the identity $\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}$ and the symmetry $\binom{2n-1}{n-1}=\binom{2n-1}{n}$

*In (4) we shift the index by one

*In (5) we use the relationship (3)

*In (6) we use the identity (1)

*In (7) we do a small rearrangement

*In (8) we use the identity (2). Since the series starts with $n=1$ we have to subtract $1$.
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