## Algebra 2 (1st Edition)

$\quad \displaystyle \sum_{n=1}^{\infty}(-2)^{n}$
The sign alternates, $(-,+,-,+,-,...)$ We use the fact that the signs of powers of $(-1)$ also alternate. $\left[\begin{array}{cccccccc} n: & 1 & 2 & 3 & 4 & ...\\ a_{n}: & (-1)^{1}\cdot 2 & (-1)^{2}\cdot 4, & (-1)^{3}\cdot 8, & (-1)^{4}\cdot 16 & \end{array}\right]$ $\Rightarrow (-1)^{n}$ is a factor of the nth term (The nth term is negative for odd n's and positive for even n's.) The absolute values of the terms are powers of 2, $\left[\begin{array}{ccccccc} n: & 1 & 2 & 3 & 4 & ...\\ a_{n}: & (-1)^{1}\cdot 2^{1} & (-1)^{2}\cdot 2^{2}, & (-1)^{3}\cdot 2^{3}, & (-1)^{4}\cdot 2^{4} & \end{array}\right]$ $\Rightarrow (2^{n})$ is a factor of the nth term $a_{n}=(-1)^{n}\cdot 2^{n}=(-2)^{n}$ The lower limit is n=1, and there is no upper limit for n. Sum = $\quad \displaystyle \sum_{n=1}^{\infty}(-2)^{n}$