# A Series For the Golden Ratio

Question: Can we show that $$\phi=\frac{1}{2}+\frac{11}{2}\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2}$$; where $$\phi={1+\sqrt{5} \above 1.5pt 2}$$ is the golden ratio ?

Some background and motivation:

Wikipedia only provides one series for the golden ratio - see also the link in the comment by @Zacky. I became curious if I could construct another series for the Golden Ratio based on a slight modification to a known series representation of the $$\sqrt{2}.$$ At first I considered $$\sqrt{2}=\sum_{n=0}^\infty(-1)^{n+1}\frac{(2n+1)!!}{(2n)!!}$$; which series can be accelerated via an Euler transform to yield $$\sqrt{2}=\sum_{n=0}^\infty(-1)^{n+1}\frac{(2n+1)!!}{2^{3n+1}(n!)^2}$$ This last series became the impetus to try and and get to the Golden ratio. Through trial and error I stumbled upon $$\frac{\sqrt{5}}{11}=\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2}$$

• $$\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}x^{n} = \frac{1}{\sqrt{1-x}}$$ (for $-1\leq x < 1$) is everything you need. – Jack D'Aurizio Dec 30 '18 at 14:59
• I am pretty sure I have seen this before on MSE. It was a post that asked if this series is true and it was painted on something like a poster, or on a book. – Zacky Dec 30 '18 at 15:20
• Fully solved here. – Did Dec 30 '18 at 20:48
• @AntonioHernandezMaquivar "Any reason why this question is being voted to [be] close[d?]" The first reason that springs to mind is the total lack of personal input, which one may find all the more surprising coming from an OP member for 3+ years and with 100+ questions asked. – Did Dec 30 '18 at 20:52
• @upvoters Any reason why this question is being upvoted? Apparently 8 users believe the question respects the etiquette of the site? Please explain. – Did Dec 30 '18 at 20:54

First of all note that

$$\frac1{\sqrt{1-4x}}=\sum_{n=0}^{\infty}\binom{2n}n x^n$$

Lets rewrite your sum as the following

$$\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2}=\frac15\sum_{n=0}^\infty\binom{2n}n\left(\frac1{5^3}\right)^n=\frac15\frac1{\sqrt{1-\left(\frac4{5^3}\right)}}=\frac{\sqrt 5}{11}$$

And therefore you can correctly conclude that

$$\frac{1}{2}+\frac{11}{2}\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2}=\frac12+\frac{11}2\frac{\sqrt 5}{11}=\frac{1+\sqrt 5}2$$

$$\therefore~\frac{1}{2}+\frac{11}{2}\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2}=\phi$$

• What is the motivation behind: $\frac1{\sqrt{1-4x}}=\sum_{n=0}^{\infty}\binom{2n}n x^n$ ? – Antonio Hernandez Maquivar Dec 30 '18 at 15:11
• Why should I have thought of $\frac1{\sqrt{1-4x}}=\sum_{n=0}^{\infty}\binom{2n}n x^n$ as a starting point to prove my own series ? – Antonio Hernandez Maquivar Dec 30 '18 at 15:16
• @AntonioHernandezMaquivar: $$\frac{d^n}{dx^n}\left.\left(\frac{1}{\sqrt{1-x}}\right)\right|_{x=0}$$ is straightforward to compute by induction or by invoking the extended binomial theorem. – Jack D'Aurizio Dec 30 '18 at 15:17
• @AntonioHernandezMaquivar Since $\frac{(2n)!}{(n!)^2}=\binom{2n}n$ which is called the central binomial coefficient. This one is used quite often for the purpose of expressing series expansions of well-know functions such as the $\arcsin$ or square root function. Thus, seeing this structure within a given series implies a possible connection to one of these functions which I conjectured and it turned out that the series expansion of the function $\frac1{\sqrt{1-4x}}$ is a good choice in this situtation. – mrtaurho Dec 30 '18 at 15:28
• You get $(1-4x)^{-1/2}=\sum_{k=0}^\infty\binom{-1/2}{k}(-4x)^k$, which is the general binomial series for $|4x|<1$. Then explore $$\binom{-1/2}{k}=\frac{(-1/2)(-1/2-1)...(-1/2-k+1)}{k!}=(-1/2)^k\frac{(2k-1)(2k-3)...3\cdot 1}{k!}=(-1/4)^k\frac{(2k)!}{(k!)^2}.$$ – LutzL Dec 31 '18 at 11:15

$$\dfrac{(2n)!}{5^{3n+1}(n!)^2}=\dfrac{2^n\cdot1\cdot3\cdot5\cdots(2n-3)(2n-1)}{5^{3n+1}n!}=\dfrac{-\dfrac12\left(-\dfrac12-1\right)\cdots\left(-\dfrac12-(n-1)\right)}{n!\cdot5}\left(-\dfrac4{5^3}\right)^n$$

$$\implies5\sum_{n=0}^\infty\dfrac{(2n)!}{5^{3n+1}(n!)^2}=\left(1-\dfrac4{5^3}\right)^{-1/2}=?$$

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

$$mx=\dfrac{2!}{5^3(1!)^2},\dfrac{m(m-1)}2x^2=\dfrac{4!}{5^6(2!)^2}$$

$$\implies m=-\dfrac12,x=-\dfrac4{5^3}$$

We know that the approximates by partial continued fractions of the golden ratios are the quotients of successive Fibonacci numbers $$1,1,2,3,5,8,13,...$$. Taking $$\phi=\frac{1+\sqrt{5}}2\approx \frac{8}5\implies \sqrt5\approx \frac{11}{5}$$ we get a lower approximation for $$\sqrt5$$. To the remainder of the square root after factoring out the approximation we can apply the the Newton/binomial series for the inverse square root $$\sqrt5=\frac{11}{5}\sqrt{\frac{125}{121}}=\frac{11}{5}\left(1-4\frac1{5^3}\right)^{-1/2} =\frac{11}5\sum_{n=0}^\infty\binom{2n}{n}\frac1{5^{3n}}$$ which is the series you got.

Exploring the same method for one place further in the Fibonacci sequence $$\phi=\frac{1+\sqrt{5}}2\approx \frac{13}8\implies \sqrt5\approx \frac{9}{4}$$ gives an upper approximation of $$\sqrt5$$ and thus an alternating series, $$\sqrt5=\frac{9}{4}\sqrt{\frac{80}{81}}=\frac{9}{4}\left(1+4\frac1{320}\right)^{-1/2} =\frac{9}4\sum_{n=0}^\infty\binom{2n}{n}\frac{(-1)^n}{320^{n}}$$

### Why that series?

The general binomial series reads as $$(1+x)^α=\sum_{n=0}^\infty\binom{α}n.$$ For $$α=-1/2$$ the binomial coefficient can be transformed as $$\binom{-1/2}{n}=\frac{(-1/2)(-1/2-1)...(-1/2-n+1)}{n!}=(-1/2)^n\frac{(2n-1)(2n-3)...3\cdot 1}{n!}=(-1/4)^n\frac{(2n)!}{(n!)^2},$$ resulting in the "simplified" formula $$\frac1{\sqrt{1-4x}}=\sum_{n=0}^\infty\binom{2n}nx^n.$$