Showing that this sequence is eventually decreasing I'm trying to show that this sequence $$a_n = \frac{3^n-7}{4^n+5}$$ is decreasing for all $n$ greater than some $N\in \Bbb N$.
All I can see to do is something like
$$a_{n+1} = \frac{3^{n+1}-7}{4^{n+1}+5} = \frac{3\cdot3^n-7}{4\cdot 4^n+5}\le \frac{3\cdot3^n-7}{4^n+5}$$
But that last expression is not less than $a_n$ for large $n$. Is there some better way to do this?

Calculate $a_{n+1}-a_n$, then
$$\frac{3^{n+1}-7}{4^{n+1}+5}-\frac{3^n-7}{4^n+5}=\frac{-3^n 4^n +21\cdot 4^n + 10\cdot 3^n}{(4^{n+1}+5)(4^n+5)}. $$
If we show $-3^n 4^n +21\cdot 4^n + 10\cdot 3^n < 0$ for large $n$, then the proof is over. Since $3^n < 4^n$ for all $n\in\mathbb{N}$,
$$-3^n 4^n +21\cdot 4^n + 10\cdot 3^n < -3^n4^n + 31\cdot 4^n=(31-3^n)4^n.$$
If $n\ge 4$, then $31-3^n < 0$, and so $(31-3^n)4^n< 0$. Therefore
$$-3^n 4^n +21\cdot 4^n + 10\cdot 3^n <(31-3^n)4^n< 0.$$
