# How many infinite series representations of the golden ratio are in existence?

How many infinite series representations of the golden ratio are in existence?

All I can find is one that expands out the $5^{1/2}$ part in $\varphi= \frac12(1+5^{1/2})$ and the one that uses the Bernoulli Numbers. Are there any more? Other numbers like $\pi$ have hundreds.

## 5 Answers

You can use that it is $\varphi = \lim_{n\to\infty} F_{n+1}/F_n$, where $F_n$ is the Fibonacci numbers to show that it: $$\varphi = 1+\frac{1}{1\cdot 1}-\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}...+\frac{(-1)^n}{F_nF_{n+1}}...$$

In some sense, $\varphi$ is the "hardest" real number to approximate with rational values. That is because the continued fraction expansion for $\varphi$ is $[1,1,1,...]$ The continued fraction expansion for a number gives a sequence of "best" rational approximations for the number in some sense, and the sequence converges faster when the coefficients are larger. Since its coefficients are all $1$, $\varphi$ has the slowest converging continued fraction expansion of all real numbers. (There are others which converge equally slowly, but none that converges slower.)

It is easy to prove geometrically, by looking at a pentagon, that

$$\cos(36^\circ) =\frac{1+\sqrt{5}}{4}$$

Thus

$$\frac{1+\sqrt{5}}{2}=2 \cos \frac{\pi}{5}$$

Using the series for $\cos(x)$ you get another representation for $the golden mean. Now square both sides, and use the double angle formula. You get another series. Repeat... In general, all$\sin$and$\cos$of multiple of$9^\circ$can be written in terms of golden mean. • Thanks for answering N.S. I though about using the taylor series for cos(x) to get a series for phi but then you end up with pi in all the terms. Unless there was a series for cos x that doesn't have pi in it's terms? – SR255 Oct 26 '12 at 15:48 • @SR255: what (sensible) series for$cos(x)$do you know that does have$\pi$in its terms? – tomasz Oct 26 '12 at 22:37 • @tomasz The Taylor series for$\cos(x)$evaluated at$\frac{\pi}{5}$;) – N. S. Oct 29 '12 at 2:43 You can find infinitely many infinite series representations for any number$x$, for instance, pick arbitrary$c$and put$x_1=(x-c)/2$,$x_n=(x+c)\cdot 2^{-n}$for$n>1$. Or$x_1=x+c,x_2=-c, x_n=0$for$n>2$. In both cases,$\sum x_n=x$. Or, if you insist on the series having rational coefficients, then pick arbitrary rational$c$and put$b_n=\lfloor nx\rfloor/n$,$x_1=-c/2+b_1$,$x_n=c\cdot 2^{-n}+b_n-b_{n-1}$for$n>1$. Then$\sum x_n=x$. I found a series $$\varphi = 3 - 2\left( \frac{\left({2\pi \over 5} \right)^2}{2!} - \frac{\left({2\pi \over 5} \right)^4}{4!} + \frac{\left({2\pi \over 5} \right)^6}{6!} - \frac{\left({2\pi \over 5} \right)^8}{8!} + ...\right)$$ • nice, is that derived from the sine function? – user58512 Jan 30 '13 at 14:06 • @user58512$1+2\cos\left(\frac{2\pi}{5}\right)\$ – Jaume Oliver Lafont Jun 3 '17 at 22:26

A family of Egyptian fractions is given by

$$\varphi = \frac{F(2n+1)}{F(2n)}-\sum_{k=0}^\infty \frac{1}{F(n2^{k+2})}$$

https://math.stackexchange.com/a/2307929/134791