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?
 A: 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 
$$\frac{a_n}{a_{n+1}}\gt \frac{4}{3}-0.1\gt 1.$$
A: 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
$$-3^n 4^n +21\cdot 4^n + 10\cdot 3^n <(31-3^n)4^n< 0.$$
A: Differentiating with respect to $n$, you get 
$(4^n + 5)(3^n  \log3 ) - (3^n - 7)(4^n  \log4)     \implies 12^n  \log(0.75)  + 5 · 3^n · \log3 + 7 · 4^n · \log4· 
\log(0.75)   <   0$;
Therefore, for large $n$, the derivative is less than $0$. 
Thus, the sequence is decreasing.
