Lower bound of $a_n=1+(-3/4)^n$

Question

I want to find the upper and lower bound of the following sequence

$$a_n=1+\left(\frac{-3}{4} \right)^n$$

For the upper bound, things are quite trivial since $\left(\frac{-3}{4} \right)^n<1$, hence $a_n\leq 2$. However, I'm getting a bit confused when it comes to the lower bound. I know that the sequence is not monotone, and therefore the first terms cannot give me a hint about what to expect.

I would appreciate if somebody could help me figure out this.

EDIT: Since we have been discussed about subsequences yet, I would prefer an answer without them.

Many thanks in advance!

Answer 1

$a_{2n} = 1 + \left(\frac{9}{16}\right)^n$ is descreasing and tends to $1$ so it stays above $1$.

$a_{2n+1} = 1 - \frac{3}{4}\left(\frac{9}{16}\right)^n$ is increasing. Its first term is $\frac{1}{4}$, so overall $$a_n \geq \frac{1}{4}$$