Determine if the sequence $\{a_n\}:a_n=(-3)^n$ is an increasing or a decreasing sequence Determine if the sequence $\{a_n\}:a_n=(-3)^n$ is an increasing or a decreasing sequence.
The $(n+1)^{st}$ term is $a_{n+1}=(-3)^{n+1}.$ So $\dfrac{a_{n+1}}{a_n}=\dfrac{(-3)^{n+1}}{(-3)^n}=\dfrac{(-3)^n(-3)^1}{(-3)^n}=-3&lt;1.$ Doesn't this mean that $a_{n+1}&lt;a_n$ for every $n$ which obviously isn't true? The sequence looks like $-3,9,-27,81...,$ so it isn't monotonic, but why the usual method tells us else? We can use it if $a_n\ne0$ for every $n$ which holds here.
It is because $\frac{a_{n+1}}{a_n} &lt; 1$ does not tell you the sequence is decreasing.
What we would all like to do is to multiply both side by $a_n$, but WE CAN'T, since $a_n$ is not necessarily positive !
So $\frac{a_{n+1}}{a_n} &lt; 1$ does not imply $a_{n+1} &lt; a_n$.
As others have remarked we cannot use your method to conclude $a_{n+1} &lt; a_{n}$. Writing out the first three terms $a_1=-3, a_2 = 9, a_3 = -27$ we see that $a_1 &lt; a_2$ which means the sequence isn't decreasing. However, we also have $a_2 &gt; a_3$ which means the sequence isn't increasing. In conclusion the sequence is neither increasing nor decreasing.