Answer
$\quad \displaystyle \sum_{n=1}^{\infty}(-2)^{n}$
Work Step by Step
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 = $\quad \displaystyle \sum_{n=1}^{\infty}(-2)^{n}$
