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Let $a,b$ be positive real number. Set $x_0 =a$ and $x_{n+1}= \frac{1}{x_n^{-1}+b}$ for $n≥0$

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

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

Any help? I don't know where to start.

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3 Answers 3

HINTS: A good place to start is with the function

$$f(x)=\frac1{x^{-1}+b}=\frac1{\frac1x+b}=\frac{x}{1+bx}\;,$$

since your recurrence is $x_{n+1}=f(x_n)$. In order to show that the sequence is monotone decreasing, you’d like to show that $f(x_n)<x_n$ for each $n\in\Bbb N$. Clearly $x_0=a>0$, and since $b>0$, it’s easy to see that $x_n>0$ for all $n\in\Bbb N$. So if $x>0$, when do we have $f(x)<x$? Clearly this happens if and only if $1+bx>1$; why, and when does that happen?

For (b), remember that a bounded monotone sequence always converges. Once you’ve done (a), you know that this sequence is monotone decreasing and bounded below by $0$, so it converges. To find the limit, use the fact that $f$ is a continuous function on the positive reals: if $L=\lim\limits_{n\to\infty}x_n$, then

$$f(L)=\lim_{n\to\infty}f(x_n)=\lim_{n\to\infty}x_{n+1}=L\;.$$

That is, $L$ is a fixed point of the function $f$. What is (or are) the fixed point(s) of $f$?

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clearly this is a sequence of positive real numbers.
Let $x_n^{-1}=k$. Then $x_{n+1}=\frac{k}{1+kb} < k$.
Hence $x_{n+1}<x_n$ and (a) follows

this is a monotonic decreasing sequence of positive real numbers hence bounded below by zero.now monotonic decreasing sequence bounded below converges to its infimum.So limit exist.I hope now you can calculate the limit.

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To prove the limit exists use the fact every decreasing bounded below sequence is convergent. To find the limit just assume $ \lim_{n\to \infty} x_n = x = \lim_{n\to \infty} x_{n+1} $ and solve the equation for $x$

$$ x=\frac{1}{1/x+b} .$$

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