1. ## Geometric Series Question

Given that the sum to infinity of a G.P. is denoted by S.

Show that if a > 0, then S > 4ar except for one value of r.

2. ## Re: Geometric Series Question

Hello, moonchildvulcan!

$\displaystyle \text{Given that the sum to infinity of a G.P. is denoted by }S.$

$\displaystyle \text{Show that if }a > 0,\,\text{ then }S > 4ar\,\text{ except for one value of }r.$

The sum of an infinite G.P. is:.$\displaystyle \frac{a}{1-r},\,\text{ for }|r| < 1.$

If $\displaystyle S$ is equal to $\displaystyle 4ar$, we have: .$\displaystyle \frac{a}{1-r} \:=\:4ar$

. . $\displaystyle \frac{1}{1-r} \:=\:4r \quad\Rightarrow\quad 1 \:=\:4r-4r^2 \quad\Rightarrow\quad 4r^2 - 4r + 1 \:=\:0$

. . $\displaystyle (2r-1)^2\:=\:0 \quad\Rightarrow\quad 2r-1 \:=\:0 \quad\Rightarrow\quad \boxed{r \:=\:\tfrac{1}{2}}$

3. ## Re: Geometric Series Question

If you start with $\displaystyle a(2r-1)^2\ge 0$ you can arrive at your conclusion.

4. ## Re: Geometric Series Question

Thanks for that. I still cannot see why S > 4ar in the first place. The rest that follows, I managed to achieve.

5. ## Re: Geometric Series Question

Originally Posted by a tutor
If you start with $\displaystyle a(2r-1)^2\ge 0$ you can arrive at your conclusion.
Sorry but why would you start with this? On what basis? It's establishing why S > 4ar in the first place that is bugging me. I am happy with solving the inequality that follows to show that r = 1/2 is the exception.

6. ## Re: Geometric Series Question

I started by assuming the conclusion to see where it lead.

$\displaystyle \frac{a}{1-r}>4ar$

$\displaystyle \Rightarrow a>4ar-4ar^2$

$\displaystyle \Rightarrow a(4r^2-4r+1)>0$

$\displaystyle \Rightarrow a(2r-1)^2>0$

but this works in reverse too so starting with $\displaystyle a(2r-1)^2 > 0$ if $\displaystyle r \ne \frac{1}{2}$ you get the result you want.