# 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.

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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.)

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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 $\sum x_n=x$.

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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 Yes you do..... – N. S. Oct 26 '12 at 16:57 @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

I found a series phi = 3 - 2{ (2/5 pi)^2/2! - (2/5 pi)^4/4! + (2/5 pi)^6/6! - (2/5 pi)^8/8! + ...

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nice, is that derived from the sine function? – user58512 Jan 30 '13 at 14:06