A golden ratio series from a comic book

Using the series $$ (1-4x)^{-1/2}=\sum_{n=0}^\infty\binom{2n}{n}x^n\tag{1} $$ we get $$ \begin{align} f(x) &=\sum_{n=0}^\infty\frac{(-1)^{n+1}(2n+1)!}{(n+2)!\,n!}x^{n+2}\\ &=\frac12\sum_{n=1}^\infty(-1)^n\binom{2n}{n}\frac{x^{n+1}}{n+1}\\ &=\frac12\int_0^x\left[(1+4t)^{-1/2}-1\right]\,\mathrm{d}t\\ &=\frac14(1+4x)^{1/2}-\frac x2-\frac14\tag{2} \end{align} $$ Therefore, $$ \begin{align} \frac{13}8+\sum_{n=0}^\infty\frac{(-1)^{n+1}(2n+1)!}{(n+2)!\,n!4^{2n+3}} &=\frac{13}8+4f\left(\frac1{16}\right)\\ &=\frac{13}8+4\left(\frac14\left(1+\frac14\right)^{1/2}-\frac1{32}-\frac14\right)\\ &=\frac{1+\sqrt5}2\\[8pt] &=\phi\tag{3} \end{align} $$

First Approach: Catalan Numbers

Some straightforward manipulations of the sum brings it to the form

$$ S=\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^{n+1}(2(n+1))!}{(n+2)!(n+1)! 4^{2n+3}} $$

Using the definition of the Catalan numbers this nicely rewrites as

$$ S=\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^{n+1}C_{n+1}}{ 4^{2n+3}}=\frac{1}{8}\sum_{n=0}^{\infty}C_{n+1}\left(\frac{-1}{16}\right)^{n+1} $$

Using the generating function of the Catalan numbers,

$$ \sum_{n=0}^{\infty}C_{n+1} x^{n+1}=\frac{1-\sqrt{1-4x}}{2x}-1 $$

setting $x=\frac{-1}{16}$ we may conclude that $$ S=\frac{1}{8} \left(4 \sqrt{5}-9\right) $$

Furthermore

$$ S+\frac{13}{8}=\frac{\sqrt{5}}{2}-\frac{9}{8}+\frac{13}{8}=\frac{1+\sqrt{5}}{2}=\phi $$

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Second Approach: Contour Integration

OK, let's see what contour integration can do.

We can show by an straightforward application of Cauchy's integral formula, that

$$ \binom{n}{k}=\frac{1}{2 \pi i}\oint_C\frac{(1+z)^{n}}{z^{k+1}}dz \quad (1) $$

where $C$ is a circle traversed counter-clockwise (with radius < $1/4$ in our case) .

Furhermore we may observe that ($C_n$ are again Catalan numbers)

$$ C_n=\frac{1}{n+1}\binom{2n}{n}=\binom{2n}{n}-\binom{2n}{n+1} \quad (2) $$

Now let's apply (1) and (2) to the function $q(x)=\sum_{n=0}^{\infty}C_{n+1}x^{n+1}$

we get

$$ q(x)=\frac{1}{2 \pi i}\oint_C\sum_{n=0}^{\infty}x^{n+1}\left(\frac{(1+z)^{2n+2}}{z^{n+2}}-\frac{(1+z)^{2n+2}}{z^{n+3}}\right)dz $$

where the exchange of summation and integration is justified as long as the poles lie not on the contour of integration (which will be the case) the sums are now usual geometric series and yield

$$ q(x)=\frac{1}{2 \pi i}\oint_C\left(-\frac{1}{z}\frac{x (z+1)^2}{x (z+1)^2-z}+\frac{1}{z^2}\frac{x (z+1)^2}{x (z+1)^2-z}\right)dz $$

It is easy to show that the zero's of the denominator are given by $z_{\pm}=\frac{-2 x\pm\sqrt{1-4 x}+1}{2 x}$ and only $z_-$ lies in the unit circle if $x<1/4$ which is important because we want to set $x=\frac{-1}{16}$ in the end. Applying the residue theorem, we get

$$ q(x)=\mathrm{res}(z=0)+\mathrm{res}(z=z_{-})=-\frac{x-1}{x}+\frac{\sqrt{1-4 x}+1}{2 x}=\\ \frac{2 x+\sqrt{1-4 x}-1}{2 x} $$

Recalling the sum you are looking for (see my first answer)is

$$ S=\frac{1}{8}q\left(\frac{-1}{16}\right) $$

we obtain

$$S+\frac{13}{8}=\phi$$ as desired.

This also constitutes a proof of the generating functions for the Catalan numbers (it is $q(x)$)!

A few easy manipulations yield the equivalent series

$$\frac{13}{8}-\frac1{64}\sum_{n=0}^\infty \frac{(2n+1)!}{n!(n+2)!}\left(-\frac1{16}\right)^n$$

The problem, then, is to evaluate

$$f(z)=\sum_{n=0}^\infty \frac{(2n+1)!}{n!(n+2)!}z^n$$

Using the duplication formula for the factorial, we have

$$\begin{align*} f(z)&=\sum_{n=0}^\infty 4^{n+\frac12}\frac{\left(n+\frac12\right)!}{\left(-\frac12\right)!}\frac{z^n}{(n+2)!}\\ &=2\sum_{n=0}^\infty \frac{\left(n+\frac12\right)!}{\left(-\frac12\right)!}\frac{(4z)^n}{(n+2)!} \end{align*}$$

We can massage the series to a more familiar form like so:

$$\begin{align*} f(z)&=\frac{2}{(4z)^2}\sum_{n=0}^\infty \frac{\left(n+\frac12\right)!}{\left(-\frac12\right)!}\frac{(4z)^{n+2}}{(n+2)!}\\ &=\frac{1}{8z^2}\sum_{n=2}^\infty \frac{\left(n-\frac32\right)!}{\left(-\frac12\right)!}\frac{(4z)^n}{n!}\\ &=-\frac{1}{4z^2}\sum_{n=2}^\infty \frac{\left(n-\frac32\right)!}{\left(-\frac32\right)!}\frac{(4z)^n}{n!}\\ &=-\frac{1}{4z^2}\sum_{n=2}^\infty \binom{n-\frac32}{n}(4z)^n\\ \end{align*}$$

We then flip the binomial coefficient to give

$$\begin{align*} f(z)&=-\frac{1}{4z^2}\sum_{n=2}^\infty \binom{\frac12}{n}(-4z)^n\\ &=\frac{1}{4z^2}\left(1-2z-\sum_{n=0}^\infty \binom{\frac12}{n}(-4z)^n\right)\\ &=\frac{1}{4z^2}\left(1-2z-\sqrt{1-4z}\right) \end{align*}$$

where the formula for the binomial series was used.

$f\left(-\frac1{16}\right)$ evaluates to $72-32\sqrt{5}$, and replacing the series with this value and simplifying finally yields $\phi$.