What is the Proof for the Sum to Infinity of an Infinite Geometric Progression?

[Appleton]

Homework Statement

An infinite geometric progression is such that the sum of all the terms after the nth is equal to twice the nth term. Show that the sum to infinity of the whole progression is three times the first term.

Homework Equations

[/B]
$S_{n} = \frac{a(1-r^n)}{1-r}\\<br /> S_{\infty} = \frac{a}{1-r}\\$
Where a is equal to the first term of the geometric series and r is equal to the common ratio.

The Attempt at a Solution

[/B]
The sum of all the terms after the nth =
$\frac{a}{1-r} - \frac{a(1-r^n)}{1-r}$

Twice the nth term =
$2ar^{n-1}\\$

So
$\frac{a}{1-r} - \frac{a(1-r^n)}{1-r} = 2ar^{n-1}\\$
I need to show that
$\frac{a}{1-r} = 3a\\$

$\frac{a}{1-r} - \frac{a(1-r^n)}{1-r} = 2ar^{n-1}\\<br /> a = 2ar^{n-1}(1-r) + a(1-r^n)\\<br /> 2ar^{n-1}=3ar^{n}\\<br /> \frac{2a}{r}=3a$

But
$\frac{2a}{r} \ne \frac{a}{1-r}$
So either I've made an error, or much less likely, the author of the book has.

 

[LCKurtz]

Homework Statement

An infinite geometric progression is such that the sum of all the terms after the nth is equal to twice the nth term. Show that the sum to infinity of the whole progression is three times the first term.

Homework Equations

[/B]
$S_{n} = \frac{a(1-r^n)}{1-r}\\<br /> S_{\infty} = \frac{a}{1-r}\\$
Where a is equal to the first term of the geometric series and r is equal to the common ratio.

The Attempt at a Solution

[/B]
The sum of all the terms after the nth =
$\frac{a}{1-r} - \frac{a(1-r^n)}{1-r}$

Twice the nth term =
$2ar^{n-1}\\$

So
$\frac{a}{1-r} - \frac{a(1-r^n)}{1-r} = 2ar^{n-1}\\$
I need to show that
$\frac{a}{1-r} = 3a\\$

$\frac{a}{1-r} - \frac{a(1-r^n)}{1-r} = 2ar^{n-1}\\<br /> a = 2ar^{n-1}(1-r) + a(1-r^n)\\<br /> 2ar^{n-1}=3ar^{n}\\<br /> \frac{2a}{r}=3a$

So either I've made an error, or much less likely, the author of the book has.

Assuming $a \ne 0$ that last equation tells you that $r = 2/3$. Use that to write down your series again and see if the result isn't true.

 

[Appleton]

Thanks for your reply, it seems I lost sight of the bigger picture.