Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
    
Compute $a_n - a_{n+1}$. Then show that the numerator is positive for large enough $n$. – Daniel Fischer Feb 6 at 15:10
up vote 8 down vote accepted

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

share|cite|improve this answer

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

share|cite|improve this answer

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.

share|cite|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.