Show that $x_n$ is increasing and $x_n < 2$ for all $n \geq 2$, for the sequence. $x_1 = 1$, $x_{n+1} = \frac{1}{5}(x_n^2 + 6)$ for $n \geq 1$.

Question

I'm doing a real analysis course at university. I'm working ahead on my issued problem sheet so there is not much documentation from my tutors regarding my question above. This is the first time I've run across induction with regard to sequences so any tips would be much appreciated.

I've tried a few times to show the sequence $\{x_n\}$ is bounded above by 2 but I'm getting lost in understanding the difference between $x_n$ and $x_{n+1}$.

Any help would be very much appreciated!

Answer 1

To show bounded above by $2$ using induction we first need to establish the base case, i.e. that $x_1<2$ (which is clear). Then we assume that $x_n<2$ and we want to show that $x_{n+1}<2$. We can do this by relating $x_{n+1}$ to $x_n$ via the definition of the sequence.

Here is an additional step if you are still stuck.

Since we're supposing that $x_n<2$ (and clearly we have $x_n>0$ by the definition), we have $x_n^2<4$, so $$x_{n+1}=\frac{x_n^2+6}{5}<...$$

Answer 2

1) Consider

$f(x)= (1/5)(x^2+6) -x$, for $0<x<2$;

$f(0)=6/5$; $f(2)=0$.

$f'(x)= (2/5)x -1 <0$ is strictly decreasing in $(0,2)$.

Hence $f(x)> 0$ in $(0,2)$.

2) Without calculus:

Need to show

$(1/5)(x^2+6) >x$ for $0<x<2;$

$f(x):= x^2-5x+6>0$;

$(x-2)(x-3)>0$;

|) $x <2$, and $x <3$, i.e $x<2$;

||) $x>2$, and $x>3$; of no interest.

$ f(x)>0$ for $x<2$, and we are done.

3) $f(x)=(1/5)(x^2+6)$;

$f'(x)=(2/5)x > 0$, for $x >0$;

$f$ is strictly increasing

Note: $x_1=1;$ $x_2=7/5,$

$x_2>x_1;$

Then $x_3= f(x_2)>f(x_1)=x_2;$

$x_4=f(x_3) >f (x_2)=x_3;$

Inductively $x_{n+1}=f(x_n) >f(x_{n-1})=x_n.$