Exact calculation of $\sum_{n=0}^\infty \cos(36^\circ)^n$

Question

Right now we are covering geometric series.To find $\cos 36^\circ$ I use the sum and difference formula.

Knowing that $\cos36^{\circ} = \frac{1+\sqrt{5}}{4} = \phi/2$ calculate $\sum_{n=0}^{\infty}\cos^n36^{\circ}$ and express it in terms of $\phi$.

Should I just use the common ratio as $\cos 36^\circ$? How can I solve this?

Knowing that $\cos18^{\circ} = \frac{-1+\sqrt{5}}{4} = \psi/2$, calculate $\sum_{n=0}^{\infty}\cos^n18^{\circ}$ and express it in terms of $\psi$.

Answer 1

Well $$S=\sum^\infty_{n=0}\cos(36º)^n=\sum^\infty_{n=0}\bigg(\frac{\varphi}2\bigg)^n$$

Now \begin{align*}S&=1+\frac{\varphi}2+\bigg(\frac\varphi2\bigg)^2+\bigg(\frac\varphi2\bigg)^3+\ldots\\\frac\varphi2S&=\;\;\quad\frac\varphi2+\bigg(\frac\varphi2\bigg)^2+\bigg(\frac\varphi2\bigg)^3+\ldots \\\end{align*}

Therefore $$S\cdot\frac{\varphi}{2}=S-1\implies S=\frac{2}{2-\varphi}$$

Use a similar approach for the other sum considering that $\psi=\varphi^{-1}$...

Answer 2

The first sum is a geometric series with ratio $\frac{\phi}{2}$ and so is equal to $$ \frac{1}{1-\frac{\phi}{2}} $$ Since $\phi^2=\phi+1$, we get $$ \frac{1}{1-\frac{\phi}{2}} = \frac{2}{2-\phi} = 2(1+\phi) $$ The second sum is handled in exactly the same way.

Answer 3

Both sums are convergent geometric series of the form $\sum_{n\ge 0}r^n=\frac{1}{1-r}$: the first is $\frac{1}{1-\phi/2}$, while the second is $\frac{1}{1-\psi/2}=\frac{1}{1-1/(2\phi)}$.