# 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 $\abs{r} <1that x < \frac{1}{2}$

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

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

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

The limiting sum of a geometric series is

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

In this case, this is

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

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

So $|-2x| < 1$

$-1 < -2x < 1$

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

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

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

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

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

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

$3 = 2(1 + 2x)$

$3 = 2 + 4x$

$1 = 4x$

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