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

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?

Answer 1

Note that if the sequence has a minimum element, it must necessarily be its infimum; similar for a maximum element.

Now the sequence has both; calculating the first $20$ elements highly indicates that $a_1 = 1$ is maximal and $a_2 = \frac{1}{31}$ is minimal. It remains to prove this fact.

Divide the sequence in the elements with odd index $O_n$ and even index $E_n$. This is to get rid of the minus sign under the exponent; a common tactic with these kind of problems. Prove that $O_n$ is decreasing and greater than $\frac{1}{3}$, while $E_n$ is increasing and smaller than $\frac{1}{3}$. The maximality/minimality of $1$ and $\frac{1}{31}$ follow.

Answer 2

$a_n=\frac{7}{3+x}-2$ 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 $\frac{1}{31}\le a_n\le 1$.