Evaluate the summation of $(-1)^k$ from $k=0$ to $k=n$ $$\sum_{k=0}^n(-1)^k$$
I know that the answer will be either -1 or 0 depending on whether there are an odd or an even number of sums in total, but how can I determine this if $k$ goes to infinity (which I am thinking means there is neither an even nor odd amount of sums).
How would I determine this?

Using the general form derived as:
\begin{align}
\sum_{k=0}^{n} x^{k} &= \sum_{k=0}^{\infty} x^{k} - \sum_{k=n+1}^{\infty} x^{k} = \frac{1}{1-x} - \frac{x^{n+1}}{1-x} = \frac{1 - x^{n+1}}{1-x}
\end{align} 
then for $x = -1$ the result becomes
$$ \sum_{k=0}^{n} (-1)^{k} = \frac{1 + (-1)^{n}}{2}.$$
