A golden ratio series from a comic book

The eighth installment of the Filipino comic series Kikomachine Komix features a peculiar series for the golden ratio in its cover:

That is,

$$\phi=\frac{13}{8}+\sum_{n=0}^\infty \frac{(-1)^{n+1}(2n+1)!}{(n+2)!n!4^{2n+3}}$$

How might this be proven?

• was there any relation to the story or they just put a cool math formula on the cover? In both cases, very cool :D
– Ant
Oct 8, 2015 at 15:34
• I don't know; I've yet to read my copy. Oct 8, 2015 at 15:38
• Probably the publisher wanted to add more fear to the story by adding a horrible formula :) Oct 20, 2015 at 7:31
• @FardadPouran: That's a good one. :) Nov 22, 2016 at 15:47
• For the sequel, they could use the next one in the family $$\phi =\frac{13}{8}+\sum_{k=0}^{\infty}(-1)^{k+1}\frac{C_{k+1}}{2^{4k+7}}$$ $$\phi =\frac{207}{2^7}+\sum_{k=0}^{\infty}(-1)^{k+2}\frac{C_{k+2}}{2^{4k+11}}$$ and so on where $C_n$ are the Catalan numbers as discussed in this post Dec 14, 2017 at 13:41

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}

• Nice! $\phantom{}$ Oct 8, 2015 at 13:27
• We have three variations on a theme... binomial series, the generating function for the Catalan numbers, and the generating function for the central binomial coefficients. If this horse isn't dead yet, perhaps we can lead it to water.
– robjohn
Oct 8, 2015 at 13:29
• The solution I previously had involved a hypergeometric route. When I noticed that the resulting hypergeometric function degenerated to the binomial series, that's when I started looking for simpler solutions... Oct 8, 2015 at 13:33
• @robjohn essentially you have given a proof of the generating function i used. Would be interesting if there are totally different approaches to this sum, maybe one could also rely on purely combinatoric arguments?! Oct 8, 2015 at 13:35
• @tired: I'd upvote again if I could. Since the numerator and denominator of $\frac{13}8$ are consecutive Fibonacci numbers, I figured that $13-8\phi$ was a power of $\phi$. Using that identity allows the elimination of that constant.
– robjohn
Oct 8, 2015 at 15:26

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 $q(x)$)!

• It's entirely your call, but I think you'd be better off splitting the contour integration approach into a different answer. Oct 8, 2015 at 15:19
• I even did this, but somehow i didn't think it's worth it... Oct 8, 2015 at 15:21

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 $\phi$.

• (If there are simpler solutions, I'd like to see them!) Oct 8, 2015 at 10:07
• Hi @J.M is back. You could identify your first sum as something like $\sim \sum_n 2\frac{(-1)^{n+1}C_{n+1}}{16^{n+1}}$ where $C_n$ are the Catalan numbers. There generating function is well known, and you can conclude easily (I think you have actually proven some kind of this approach). Oct 8, 2015 at 10:12
• If you are interested, i could provide another proof using contour integration later/tomorrow Oct 8, 2015 at 14:11
• That would be nice to see! Oct 8, 2015 at 14:18
• Done, hope you enjoy it! Oct 8, 2015 at 15:10

This series appears in this place, referencing the derivation at "Golden Mean Series Derivation", both by Brian Roselle. That derivation starts with $$\phi = \frac{\sqrt{5}+1}{2},$$ replaces $\sqrt{5}$ with $f(x) = \sqrt{x},$ expands $f$ in a Taylor series about $4$ (having a radius of convergence $> 1$, although this is not stated), then evaluating that series at $5$. Mr. Roselle describes finding the form of $f^{(a)}(x)$ via inspection.

We, however, can find these via induction. We wish to find $f^{(a)}(4)$:

• $f^{(0)}(4) = 4^\frac{1}{2} = 2$.
• $f'(4) = \frac{1}{2} 4^\frac{-1}{2} = \frac{1}{4}$.
• $f''(4) = \frac{1}{2} \frac{-1}{2} 4^\frac{-3}{2} = \frac{-1}{32}$.
• Suppose $f^{(n)}(4) = \left. \frac{1}{2} \cdot\left( \frac{-1}{2} \cdot \frac{-3}{2} \cdot \cdots \cdot \frac{3-2n}{2} \right) x^{\frac{1}{2}-n} \right|_{x=4}$, which is certainly true for $n=2$. Then $\left(f^{(n)}\right)'(4) = \left. \frac{1}{2} \cdot\left( \frac{-1}{2} \cdot \frac{-3}{2} \cdot \cdots \cdot \left( \frac{3}{2}-n \right) \cdot \left( \frac{3}{2}-(n+1) \right) \right) x^{\frac{1}{2}-(n+1)} \right|_{x=4} = f^{(n+1)}(4)$.

So induction has given us, for $n \geq 2$, \begin{align} f^{(n)}(4) &= \left. \frac{1}{2} \cdot\left( \frac{-1}{2} \cdot \frac{-3}{2} \cdot \cdots \cdot \frac{3-2n}{2} \right) x^{\frac{1}{2}-n} \right|_{x=4} \\ &= 2^{-n}(-1)^{n-1}(2n-3)!! 4^{\frac{1}{2}-n}, \end{align} where we have used $k!! = \begin{cases} k(k-2)(k-4) \cdots 4 \cdot 2, & k \text{ even}, \\ k(k-2)(k-4) \cdots 3 \cdot 1, & k \text{ odd} \end{cases}$. Then for $x$ close enough to $4$ (and a quick ratio test shows $5$ is close enough), \begin{align} f(x) &= \sum_{n=0}^\infty \frac{f^{(n)}(4)}{n!}(x-4)^n \\ &= \sum_{n=0}^\infty \frac{2^{-n}(-1)^{n-1}(2n-3)!! 4^{\frac{1}{2}-n}}{n!}(x-4)^n. \end{align} Replace $(2n-3)!!$ with $\frac{(2n-3)!}{2^{n-2}(n-2)!}$ and $x-4$ with $1$, pull out the first two summands and reindex $\sum_{n=2}^\infty \dots$ to start at $n=0$, plug into $\frac{f(5)+1}{2}$, and you get the given form of $\phi$.

• Very interesting. It could be that the comic's author did get the series from there. Oct 9, 2015 at 4:28

$$T(n)=\frac{(-1)^{n+1}(2n+1)!}{(n+2)!n!4^{2n+3}}=\dfrac1{64}\cdot\dfrac{1\cdot3\cdot5\cdots(2n+1)2^n}{(n+2)!16^n}$$

As the denominator has $$(n+2)!,$$

$$T(n)=-\dfrac1{8^{n+2}}\cdot\dfrac{-1\cdot1\cdot3\cdot5\cdots(2n+1)}{(n+2)!}$$ $$=\dfrac1{4^{n+2}}\cdot\dfrac{\dfrac12\left(\dfrac12-1\right)\left(\dfrac12-2\right)\cdots\left(\dfrac12-(n+1)\right)}{(n+2)!}$$

Using Generalized Binomial Expansion, $$(1+x)^m=1+mx+\frac{m(m-1)}{2!}x^2+\frac{m(m-1)(m-2)}{3!}x^3+\cdots$$ given the converge holds

$$\sum_{n=0}^\infty T(n)=\left(1+\dfrac14\right)^{1/2}-1-T(-1)=\dfrac{\sqrt5}2-1-\dfrac14\dfrac12$$