How to prove that $\sum_{n=1}^{\infty} \frac{\phi^{n}-1}{\phi^{2n}} = 1$?

Question

How can one prove that $$\sum_{n=1}^{\infty} \frac{\phi^{n}-1}{\phi^{2n}} = 1$$ where $\phi$ is the golden ratio?

Answer 1

As $\dfrac1\phi<1,$

$$\sum_{n=1}^\infty\left(\dfrac{\phi^n-1}{\phi^{2n}}\right)$$

$$=\sum_{n=0}^\infty\left(\dfrac{\phi^n-1}{\phi^{2n}}\right)$$

$$=\sum_{n=0}^\infty\left(\dfrac1\phi\right)^n-\sum_{n=0}^\infty\left(\dfrac1\phi\right)^{2n}$$

$$=\dfrac1{1-\dfrac1\phi}-\dfrac1{1-\dfrac1{\phi^2}}$$

Now $\phi$ satisfies $t^2-t-1=0$

Answer 2

The sum in question $\sum_{n=1}^{\infty} \frac{\phi^{n}-1}{\phi^{2n}} = 1$

you can write

$$\sum_{n=1}^{\infty} \frac{1}{\phi^{2n-1}}$$

which is a geometric progression whose first term is $a_1=\frac{1}{\phi}$ and the common ratio is equal to $q=\frac{1}{\phi^{2}}$.

The infinite sum is equal to $\frac{a_{1}}{1-q}$ replacing you have: $\frac{\phi}{\phi^{2}-1}$ which is equal to 1.