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