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?

We have
$$\frac{a_{n}}{a_{n+1}}=\frac{3^n-7}{3^{n+1}-7}\cdot \frac{4^{n+1}+5}{4^n+5}.$$
The ratio $\dfrac{a_n}{a_{n+1}}$ has limit $4/3$ as $n\to\infty$. (Divide top and bottom of the first term on the right by $3^n$, and top and bottom of the second term by $4^n$.)
So by the definition of limit there is an $N$ such that if $n\gt N$ then 
$$\frac{a_n}{a_{n+1}}\gt \frac{4}{3}-0.1\gt 1.$$
