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

Question

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.

Answer 1

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.

Answer 2

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 $\sum x_n=x$.

Answer 3

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.)

Answer 4

I found a series $$\varphi = 3 - 2\left( \frac{\left({2\pi \over 5} \right)^2}{2!} - \frac{\left({2\pi \over 5} \right)^4}{4!} + \frac{\left({2\pi \over 5} \right)^6}{6!} - \frac{\left({2\pi \over 5} \right)^8}{8!} + ...\right)$$