Find the sum of golden ratios in geometric series

Question

First: Do i need to determine whether these sums converge or diverge?

My attempt:

1. $\sum_{n=0}^{\infty} (\frac{1}{\phi})^n$

$r = \frac{1}{\phi}$

sum is then $\frac{1}{1-\frac{1}{\phi}}$

2. $\sum_{n=0}^{\infty} (\frac{1}{\phi})^{2n}$

$r = \frac{1}{\phi}$

sum is then $\frac{1}{1-\frac{1}{\phi}}$

3. $\sum_{n=0}^{\infty} (\frac{1}{\phi})^{2n+1}$

I have the same answer as the previous two problems.

I feel like I am missing something or doing something wrong, can someone please help?

Answer 1

You've made a slight error on $2)$ and $3)$. For instance, the limit in $2)$ should be $\dfrac1{1-(\dfrac1\phi)^2}$.

For $3)$, try factoring out $\dfrac1\phi$.

Answer 2

In general, the formula for an infinite geometric sum is

$$\sum_{k=0}^\infty r^k = \frac{1}{1-r}$$

If there is a first term $a$ then we can also say

$$\sum_{k=0}^\infty ar^k = \frac{a}{1-r}$$

In this light...

• The first series has first term $a=1$ and ratio $r = 1/\phi$.
• The second series has first term $a=1$ and ratio $r = 1/\phi^2$.
• The third series has first term $a=1/\phi$ and ratio $r = 1/\phi^2$. (Equivalently, you could factor out $1/\phi$ first and then have a series of the same ratio but first term $a=1$. Then you could find the sum's value and multiply back through by $1/\phi$.)

Plug in the corresponding values into the formula, and simplify as necessary.

Your values for the second and third series are incorrect, because you're skipping terms in the series, yet the formula you're applying doesn't permit that. It might prove useful to see how close you are to having the right formulas for them if you take note of what I mentioned about the first term earlier, and then notice that

$$x^{yz} = (x^y)^z$$

for nonnegative reals $x,y,z$.

Also, regarding convergence, technically you should check it convergences, but it is easy to show. A geometric series convergences for $|r|<1$ and $r=1/\phi$ satisfies that as

$$0< \frac{1}{\phi} = \frac{2}{1 + \sqrt 5} < \frac{2}{1 + \sqrt 4} = \frac{2}{3} < 1$$

Answer 3

You have the right idea of starting with the geometric series formula for the infinite sum $S = \frac a{1-r}$ (where $a$ is the first term and $r$ is the common ratio), but there are two things you should do to improve your results:

1) first prove convergence by establishing that $|r| < 1$, which shouldn't be too tough in each case (note that $0 < \frac 1{\phi^2}< \frac 1{\phi} < 1$).

2) simplify by noting that $\phi^2 - \phi - 1 = 0$ which you can have loads of fun rearranging.

So, for the first one ($a = 1, r = \frac 1{\phi}$), the sum is $\frac 1{1 - \frac 1{\phi}} = \frac{\phi}{\phi - 1} = \frac{\phi^2}{\phi^2 - \phi} = \phi^2 = \phi + 1$

The second ($a = 1, r = \frac 1{\phi^2}$) is $\frac 1{1-\frac 1{\phi^2}} = \frac{\phi^2}{\phi^2 - 1} = \frac{\phi^2}{\phi} = \phi$

And the third ($a = \frac 1{\phi}, r = \frac 1{\phi^2}$) is $\frac{\frac 1\phi}{1 - \frac 1{\phi^2}} = \frac{\phi}{\phi^2-1} = \frac{\phi}{\phi} = 1$