TITLE: How to determine the supremum and infimum of $a_n = \frac{(-2)^{n+1} + 3^{n}}{3^{n+1} + (-2)^n}$ for $n \in \mathbb{N}$?
QUESTION [1 upvotes]: I have successfully proven that $\lim(a_n) = \frac{1}{3}$ by shortening the fraction by $3^{n+1}$ and determining the limit of the resulting term. From this, I conclude that $a_n$ is bounded.
Now I want to find the infimum and supremum of this sequence. Judging the first few elements, I suspect that they are $\frac{1}{3}$ and $\frac{43}{73}$ respectively. However, I can't find a way to prove it. What approach would you take here?

REPLY [3 votes]: $a_n=\frac{7}{3+x}-2$ where $x=(-\frac{2}{3})^{n}$.
$-\frac{2}{3}\le x\le\frac{4}{9}$, if $n=0$ is not included.
Therefore  $\frac{7}{3}\le 3+x\le \frac{31}{9}$ 
and $\frac{1}{31}\le a_n\le 1$.
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