Manipulating a version of summations to produce the closed formula for the Catalan Numbers. Manipulate $g(x) = \frac{1}{2x}\left(1 - \sum_{k=0}^{\infty}{1/2 \choose k}(-4x)^k\right)$ into $g(x) = \sum_{n=0}^{\infty}{1/2 \choose n+1}(-1)^n2^{2n+1}x^n$
I know that I should substitute n + 1 for k to allow the summation to begin at 0, so I can see where part of the second representation comes from, but am lost on how to manipulate the other parts to obtain the desired form. Can anyone explain how to do so?
 A: $$\begin{align*}
\frac1{2x}\left(1-\sum_{k\ge 0}\binom{1/2}k(-4x)^k\right)&=\frac1{2x}\left(1-\sum_{k\ge 0}\binom{1/2}k(-1)^k2^{2k}x^k\right)\\
&=\frac1{2x}\left(1-\sum_{k\ge -1}\binom{1/2}{k+1}(-1)^{k+1}2^{2k+2}x^{k+1}\right)\\
&=\frac1{2x}-\sum_{k\ge -1}\binom{1/2}{k+1}(-1)^{k+1}2^{2k+1}x^k\\
&=\frac1{2x}+\sum_{k\ge -1}\binom{1/2}{k+1}(-1)^k2^{2k+1}x^k\\
&=\frac1{2x}+\binom{1/2}0(-1)^{-1}2^{-1}x^{-1}+\sum_{k\ge 0}\binom{1/2}{k+1}(-1)^k2^{2k+1}x^k\\
&=\frac1{2x}-\frac1{2x}+\sum_{k\ge 0}\binom{1/2}{k+1}(-1)^k2^{2k+1}x^k\\
&=\sum_{k\ge 0}\binom{1/2}{k+1}(-1)^k2^{2k+1}x^k
\end{align*}$$
A: 
A slightly different variation:
\begin{align*}
\frac{1}{2x}\left(1-\sum_{k=0}^\infty\binom{\frac{1}{2}}{k}(-4x)^k\right)&=
\frac{1}{2x}\left(-\sum_{k=1}^\infty\binom{\frac{1}{2}}{k}(-4x)^k\right)\tag{1}\\
&=\frac{1}{2x}\left(-\sum_{k=1}^\infty\binom{\frac{1}{2}}{k}(-1)^k2^{2k}x^k\right)\tag{2}\\
&=\sum_{k=1}^\infty\binom{\frac{1}{2}}{k}(-1)^{k-1}2^{2k-1}x^{k-1}\tag{3}\\
&=\sum_{k=0}^\infty\binom{\frac{1}{2}}{k+1}(-1)^{k}2^{2k+1}x^{k}\tag{4}\\
\end{align*}

Comment:


*

*In (1) the summand with $k=0$ cancels out

*In (2) we write $(-4x)^k=(-1)^k2^{2k}x^k$

*In (3) we multiply each summand of the series with $-\frac{1}{2x}$

*In (4) we shift the index $k$ by one to start from zero
