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Let the sequence $\{a_n\}$ be defined as $a_1=1$ and $$a_{n+1} = \frac{6a_n+3}{a_n+4}$$

Then I'm asked to show :

$1)$ $a_n \lt 3$.

$2)$ Assuming $a_n \lt 3$, show that the sequence is increasing.

For the first part I tried to use Induction. When I assumed $a_n \lt 3$ and went on to prove that this implies $a_{n+1} \lt 3$, I got stuck : $$ a_{n+1} = \frac{6a_n+3}{a_n+4} \lt \frac{6\cdot 3+3}{a_n+4} \lt \frac{21}{a_n+4}.$$

I'm stuck here can anyone please help?

Edit : I can do the second part easily. And also I got an appropriate answer by now for the first part.

Sathvik
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Itachi
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  • You could also show $f(x)=\frac{6x+3}{x+4}$ maps $[1,3]$ to $[f(1),f(3)]$ by showing $f$ is increasing and continuous on $[1,3]$. Once you do this $(a)$ falls right out. – Matthew H. Nov 27 '21 at 15:56

3 Answers3

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Observe,$$a_{n+1}=\frac{3(a_{n}+4)-(9-3a_n)}{a_n+4}=3-\frac{9-3a_n}{a_n+4}$$ If $a_n<3$, $$\dfrac{9-3a_n}{a_n+4}>0 \implies a_{n+1}<3$$

Sathvik
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If $a_n \lt 3$, then

$$a_{n+1}= \frac {6a_n+3}{a_n+4}=\frac{6a_n+24-21}{a_n+4}=6-\frac{21}{a_n+4} \lt 6-\frac{21}{7}=3.$$

Robert Shore
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Just for your curiosity.

If you look at this question, I described the steps for a first-order rational difference equation. If you just follow the steps, your coefficients being nice, you should end with $$a_n=\frac{9\ 7^n-7\ 3^n}{7\ 3^n+3\ 7^n}=3-\frac{28\ 3^n}{7\ 3^n+3\ 7^n}$$

Claude Leibovici
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