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Let $a,b$ be positive real number. Set $x_0 = a$ and $x_{n+1} = 1/[(1/x_n) + b]$ for $n ≥ 0$

(a) Prove that $x_n$ is monotone decreasing.

(b) Prove that the limit exists and find it.

My work:

(a) By given premises, $x_n ≥ 0$ for each n,

$x_1 = x_{0+1} = 1/[(1/x_0) + b] = 1/[(1/a) + b]$

$x_2 = x_{1+1} = 1/[(1/x_1) + b] = 1/[(1/a) + b + b] = 1/[(1/a) + 2b]$

and then continues.

we can see as $n$ increases, $x$ dereases as denominator increases.

$n→∞, nb→∞$, the sequences decreases

(then i dont know how to continue with induction)

(b) By Monotone Convergence Theorem, the sequence ($x_n$) is convergent, bounded below, limit exists

....But I dont know how to find that limit ....

Thank you guys!!Please

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What did you mean to say with that L there in the first line? –  DonAntonio Jan 30 '13 at 20:24
    
oh.. Im sorry I cant really get why $1/(1/a + b)=1/(a+b)$ –  Paul Jan 30 '13 at 20:28
    
However, isnt that enough as I showed in the denominator $1/a$ will always keep constant as $n$ increases and $b$ will always be multiply by a positive natural coefficient such that denominator is getting bigger and bigger? –  Paul Jan 30 '13 at 20:38
2  
@DonAntonio The sequence is indeed decreasing. Your calculation of $x_1$ is wrong. It should be $a/(1+ab)$. –  1015 Jan 30 '13 at 20:51
    
Ok, dear...hehe. Thanks for that. –  DonAntonio Jan 30 '13 at 21:32

1 Answer 1

up vote 1 down vote accepted

Let us compute: $$ x-f(x)=x-\frac{x}{1+bx}=\frac{bx^2}{1+bx}. $$ So we see that $x-f(x)>0$ for all $x>0$.

An easy induction proof shows that $x_n>0$ for all $n$.

So for all $n$, $$ x_n-f(x_n)=x_n-x_{n+1}>0. $$

Hencee, indeed, the sequence $(x_n)$ is decreasing and bounded below by $0$. So it converges to some limit $L\geq 0$.

Now since $f$ is continuous, we must have at the limit $$ L=f(L)\quad\Leftrightarrow\quad L=\frac{L}{1+bL}\quad\Leftrightarrow\quad L=0. $$

So $(x_n)$ converges to $0$.

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hi, how come $x_n ≤ x_0 = a$ will result in $1/(1/x_{n-1}+b) ≥ 1/(1/a +b)$ ??? –  Paul Jan 30 '13 at 21:01
    
So we have $$x_{n+1}=\frac{x_n}{1+bx_n}$$ and assuming $\,x_n\xrightarrow [n\to\infty]{}L\neq 0\,$ , we get by arithmetic of limits $$L=\frac{L}{1+bL}\Longrightarrow1+bL=1\Longrightarrow bL=0\ldots$$ In fact, @Julien, I think that in your answer we should have $\,L=f(L)=\frac{L}{1+bL}\,$ , just as is written above. –  DonAntonio Jan 30 '13 at 21:38
    
Thank you both! –  Paul Jan 30 '13 at 21:46
    
@DonAntonio Thanks! Edited. –  1015 Jan 30 '13 at 22:14
    
@Dominic I was wrong. Now it should be fine. –  1015 Jan 30 '13 at 22:15

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