Answer
The absolute values of the ratios are not smaller than $1$
Work Step by Step
Remember, the geometric series
\begin{equation}
\sum_{k=0}^\infty ar^k
\end{equation}
converges if $|r|<1$ and diverges if $|r|\geq1$. If the series converges, then the sum is
\begin{equation}
\sum_{k=0}^\infty ar^k = \frac{a}{1-r}
\end{equation}
For $r=x$, $a=1$ and $|x|<1$ we have
\begin{equation}
\frac{1}{1-x} = 1 + x + x^2 + x^3 + \dotsb = \sum_{n=0}^\infty x^n
\end{equation}
Thus, the formula doesn't hold true for $x=-1$ and $x=2$ because $|x|\not <1$.
