# Thread: appeal to the geometric series

1. ## appeal to the geometric series

Derive the indicated result by appealing to the geometric series

$\sum\limits_{k = 0}^\infty$ $(-1)^kx^k =1/(1+x),$ |x| < 1

$\sum\limits_{k = 0}^\infty$ $(-1)^kx^{2k} =1/(1+x^2),$ |x| < 1

I dun understand what the question is asking, the geometric series is 1/(1-x), how does it turn into 1/(1+x)

2. Originally Posted by wopashui
Derive the indicated result by appealing to the geometric series

$\sum\limits_{k = 0}^\infty$ $(-1)^kx^k =1/(1+x),$ |x| < 1

$\sum\limits_{k = 0}^\infty$ $(-1)^kx^{2k} =1/(1+x^2),$ |x| < 1

I dun understand what the question is asking, the geometric series is 1/(1-x), how does it turn into 1/(1+x)
You should know that $\sum\limits_{k = 0}^{\infty} r^k = \frac{1}{1 - r}$ for $|r| < 1$.

In your first question $r = -x$ and in your second question $r = -x^2$.

3. Originally Posted by wopashui
I dun understand what the question is asking, the geometric series is 1/(1-x), how does it turn into 1/(1+x)
Simple. Just replace x by –x.