# 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?

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Compute $a_n - a_{n+1}$. Then show that the numerator is positive for large enough $n$. – Daniel Fischer Feb 6 at 15:10

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 $$\frac{a_n}{a_{n+1}}\gt \frac{4}{3}-0.1\gt 1.$$

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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 $$-3^n 4^n +21\cdot 4^n + 10\cdot 3^n <(31-3^n)4^n< 0.$$

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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.

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I am not sure ! $a_0=-1$, $a_1=-\frac{4}{9}$, $a_2=\frac{2}{21}$, $a_3=\frac{20}{69}$ are increasing. Starting from $a_4=\frac{74}{261}$ the terms decrease. How do you prove that the derivative cancels somewhere around $n=3.38$ if we treat $n$ as a continuous variable ? – Claude Leibovici Feb 6 at 15:41
@ClaudeLeibovici, If you really wanna prove that, you can differentiate again. It would be easy then. The question here is for large numbers. for large numbers, there is no problem. – Win Vineeth Feb 6 at 23:52