# Thread: [SOLVED] Limiting Sum of Geometric Series

1. ## [SOLVED] Limiting Sum of Geometric Series

For what range of values of x will the series 1-2x+4x^2-8x^3+... have a limting sum? What is the value of x if the limting sum is 2/3?

Is it because $\displaystyle \abs{r} <1that x < \frac{1}{2}$

2. Originally Posted by xwrathbringerx
For what range of values of x will the series 1-2x+4x^2-8x^3+... have a limting sum? What is the value of x if the limting sum is 2/3?

Is it because $\displaystyle \abs{r} <1that x < \frac{1}{2}$
Writing this series in sigma notation we have

$\displaystyle 1 - 2x + 4x^2 - 8x^3 + \dots = \sum_{i = 0}^{\infty}(-2x)^i$.

This is a geometric series with $\displaystyle a = 1$ and $\displaystyle r = -2x$.

The limiting sum of a geometric series is

$\displaystyle S_{\infty} = \frac{a}{1 - r}$ provided that $\displaystyle |r| < 1$

In this case, this is

$\displaystyle \frac{1}{1 - (-2x)} = \frac{1}{1 + 2x}$.

But of course, this is provided that $\displaystyle |r| < 1$

So $\displaystyle |-2x| < 1$

$\displaystyle -1 < -2x < 1$

$\displaystyle \frac{1}{2} > x > -\frac{1}{2}$

$\displaystyle -\frac{1}{2} < x < \frac{1}{2}$

$\displaystyle |x| < \frac{1}{2}$.

So the limiting sum is $\displaystyle \frac{1}{1 + 2x}$ for $\displaystyle |x| < \frac{1}{2}$.

If the limiting sum is $\displaystyle \frac{2}{3}$ then

$\displaystyle \frac{1}{1 + 2x} = \frac{2}{3}$

$\displaystyle 3 = 2(1 + 2x)$

$\displaystyle 3 = 2 + 4x$

$\displaystyle 1 = 4x$

$\displaystyle x = \frac{1}{4}$.