proof by induction that $a_n$ < 5, when $ a_{n+1} = \frac{6a_n + 5}{a_n + 2} $

Question

$ a_1 = 1, $

$ a_{n+1} = \frac{6a_n + 5}{a_n + 2} $ (where n is a +ve integer)

we have to prove by induction that $ a_n $ is always less than 5.

so I used induction to prove that $ a_n $ is always positive.

And i calculated $ a_2 = \frac{11}{3} $ for the base case

then we assume that $ a_k < 5 $

so $ 6a_k + 5 < 35 $ and $ a_k + 2 > 2 $

so $ a_{n+1} = \frac{6a_n + 5}{a_n + 2} < \frac{35}{2} = 17.5 $ --> equation 1

so I figured if i could get a bigger value for the denominator, this approach would work. so i tried calculating the biggest value $ a_k $ can take (i drew a graph) , but it turns out that it is 5 (what i mean is that while $ a_n < 5, a_{n+1} > a_n $, but when $ a_n > 5 $ then the value of $ a_{n+1} < a_n $

However, this doesn't prove $a_n < 5 $, it proves that $ a_n <= 5 $

so I just substitute this in equation 1, instead of 35/2 write 35/7 and get the result

finally, my question is, is there an easier way to do this? a major part of my result relies on the graph,, so is there any way i could do it using induction only?

Answer 1

You have $0<a_n<5$.

Then $a_{n+1}=6-\frac{7}{a_n+2}<6-\frac{7}{7}=5$ which is exactly what you need.

Answer 2

Hint:

The homographic function $\:f(x)=\dfrac{6x+5}{x+2}$ is continuous, increasing on the interval $(-2+\infty)$, so $$f\bigl([0,5]\bigr)=\bigl[f(0),f(5)\bigr]=[3,5].$$

Answer 3

Assuming $x>0$ (so $x+2>0$),

$x<5\iff 6x+5<5x+10\iff \dfrac{6x+5}{x+2}<5$