Show a sequence is decreasing

**hockey777** · September 23rd 2008, 06:06 PM

$x_{n+1}=2-\frac{1}{x_n}$

I'm trying to do this by induction, I can actually show this for the case k=n. I cannot however do it for the case =1.

Also $x_1>1$

Can anyone help me show $x_1$ > $x_2$ ?

**hockey777** · September 24th 2008, 02:41 AM

Sorry, I was editing while you were responding.

$x_1>1$

**hockey777** · September 24th 2008, 06:17 AM

Actually I just need to show it's
Monotone

**bkarpuz** · September 24th 2008, 07:10 AM

Originally Posted by hockey777

$x_{n+1}=2-\frac{1}{x_n}$

I'm trying to do this by induction, I can actually show this for the case k=n. I cannot however do it for the case =1.

Also $x_1>1$

Can anyone help me show $x_1$ > $x_2$ ?

Section A
Let $x$ be nonincreasing and $x(1)>1$ .
Then, we have
$x(2)=2-\frac{1}{x(1)}=1+1-\frac{1}{x(1)}>1+\frac{1}{x(1)}-\frac{1}{x(1)}=1,$
which indicates by induction that $x(n)>1$ for all $n\in\mathbb{N}$ .
Clearly, $\ell:=\lim\nolimits_{n\to\infty}x(n)$ exists and $\ell\geq1$ holds.
Passing limit as $n\to\infty$ on both sides of the recursion relation, we get $\ell=2-\frac{1}{\ell}$ or equivalently $(\ell-1)^{2}=0$ .
Thus, $\ell=1$ must be true.

However, I really do not think that the given relation and the intitial condition gives $x(2)>x(1)$ .

Section B
But if you can show $x(2)<x(1)$ , the rest follows by induction since
$x(n+1)-x(n)=\frac{x(n)-x(n-1)}{x(n)x(n-1)}>0$
is true if $x(n)-x(n-1)>0$ holds.

I will be thinking on it...

Section C
Set $f(t):=\frac{1}{t}+t$ for $t\in(1,\infty)$ .
Hence, it is easy to see that $f(t)>2$ for all $t\in(1,\infty)$ since $f^{\prime}(t)=1-\frac{1}{t^{2}}>0$ holds for all $t\in(1,\infty)$ .
Now, we shall show $x(2)-x(1)<0$ , which implies $2-\frac{1}{x(1)}-x(1)=2-f(x(1))<0$ .
Apply Section B to obtain the result.

Section D
Therefore, $x$ is a decreasing sequence which tends to $1$ from above at $\infty$ .

**hockey777** · September 24th 2008, 07:30 AM

^To complicated, but got me on the right track

$(x_{1}-1)^{2}>0$

$x_{1}^{2}>2x_1-1$

$x_1>\frac{2x_1-1}{x_1}$

$x_1>2-\frac{1}{x_1}$

$x_1>x_2$

**hockey777** · September 24th 2008, 09:29 AM

Could someone help me show that 1 is the infinum of this set?

**bkarpuz** · September 24th 2008, 09:32 AM

Originally Posted by hockey777

Could someone help me show that 1 is the infinum of this set?

I have it proven in my post, read it carefully.

**hockey777** · September 24th 2008, 10:25 AM

I saw that, but it's not really the approach I'm trying to master right now.

Does anyone else have anything else they can recommend?

**bkarpuz** · September 24th 2008, 10:27 AM

Originally Posted by hockey777

I saw that, but it's not really the approach I'm trying to master right now.

Does anyone else have anything else they can recommend?

You know its decreasing now, this means that your sequence has limit at infinity, and this limit value is exactly infimum of this set.

What way you want to show it?
Give me some clues?

**hockey777** · September 24th 2008, 10:32 AM

The question asks to prove it's monotone and convergent, then to find the limit.

I've proven it's monotone and convergent; by the Monotone Convergence Theorem I can show that the lim= inf( $x_n$ ).

What you did in your original post isn't tools we've used yet to prove infinums of sets yet. Hence, I've got to figure out another way to do it, and every way I"ve tried has ended in failure.

**bkarpuz** · September 24th 2008, 10:48 AM

Originally Posted by hockey777

The question asks to prove it's monotone and convergent, then to find the limit.

I've proven it's monotone and convergent; by the Monotone Convergence Theorem I can show that the lim= inf( $x_n$ ).

What you did in your original post isn't tools we've used yet to prove infinums of sets yet. Hence, I've got to figure out another way to do it, and every way I"ve tried has ended in failure.

First, we show that its decreasing, next we show that it is bounded from below (in worst case it is bounded below by $0$ ), this means that it has a limit at infinity.
Then in the recursion we pass to limits on both sides (note that if $x(n)\to\ell$ then $x(n+1)\to\ell$ as $n\to\infty$ ; that, is every subsequence converges to the same limit).
Finally, we prove that $\ell=1$ , which is the desired $\inf$ value since $x$ is decreasing.

I know I repeat the same, but no other idea to show $\ell=1$ .

**Plato** · September 24th 2008, 11:36 AM

Originally Posted by bkarpuz

First, we show that its decreasing, next we show that it is bounded from below (in worst case it is bounded below by $0$ ), this means that it has a limit at infinity.
Then in the recursion we pass to limits on both sides (note that if $x(n)\to\ell$ then $x(n+1)\to\ell$ as $n\to\infty$ ; that, is every subsequence converges to the same limit).
Finally, we prove that $\ell=1$ , which is the desired $\inf$ value since $x$ is decreasing.
I know I repeat the same, but no other idea to show $\ell=1$ .

Hockey777 the above is correct. It is the standard way to work this problem.
BTW: Did you know that $\forall a > 1\quad \Rightarrow \quad a + \frac{1}{a} > 2$ ?
That is the quick way to work this problem using induction.

**bkarpuz** · September 24th 2008, 11:40 AM

Originally Posted by Plato

Hockey777 the above is correct. It is the standard way to work this problem.
BTW: Did you know that $\forall a > 1\quad \Rightarrow \quad a + \frac{1}{a} > 2$ ?
That is the quick way to work this problem using induction.

This is just proved in my first post (See Section C in my 1st post).

**hockey777** · September 24th 2008, 11:59 AM

i figured it out and got it, thanks guys

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