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Here's my question to prove:

Define $a_n$ to be a sequence such that:

$$a_1=\frac{3}{2}$$ $$a_{n+1}=3-\frac{2}{a_n}$$

Prove that $a_n$ is convergent and calculate its limit.

Solution

Prove by induction that $a_n$ is monotonic increasing:

  • For $n=2$, $a_2=3-\frac{4}{3}=\frac{5}{3}>\frac{1}{2}$
  • Assume that $a_n>a_{n-1}$
  • For $n=k+1$: $$3-\frac{2}{a_n} - (3-\frac{2}{a_{n-1}})=-\frac{2}{a_n}+\frac{2}{a_{n-1}}>0$$, since $a_{n-1}<a_n$ which makes $\frac{2}{a_{n-1}}>\frac{2}{a_n}$

  • Therefore, the sequence is monotonic increasing.

Prove that $2$ is an upper bound of the sequence. Therefore, it is monotonic increasing and bounded, thus convergent (induction).

Now I think that $\lim_\limits{n\to\infty} a_n=2$, and I want to prove it with the squeeze theorem.

Is my solution, correct?

Is there a way to find supremum here?

Thanks,

Alan

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  • $\begingroup$ Have you shown that $2$ is an upper bound? Assuming you have, then let $L$ be the limit. Then $L=3-\frac 2L$. $\endgroup$ – lulu Sep 5 '15 at 19:05
  • $\begingroup$ @lulu I think I have, by induction. With your method I am just solving a quadratic equation? Why is it "legal" to do it? how do I just "decide" that 2=$L$? $\endgroup$ – Alan Sep 5 '15 at 19:07
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    $\begingroup$ If so, then you have shown that the limit $L$ exists, do you see why? Then write $a_{n+1} = 3- \frac {2}{a_n}$ and take limits of both sides, invoking the continuity of the function $\frac 1x$. $\endgroup$ – lulu Sep 5 '15 at 19:09
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To prove that $2$ is an upper bound of the sequence, use the fact that $a_{n}$ is monotone increasing and $a_{1}=3/2$ so that $a_{n}\geq3/2$, or $1/a_{n}\leq2/3$.

The limit of the sequence is a fixed point of the function $x\mapsto3-2/x$. You can find the fixed points of this function by solving $x=3-2/x$ (hint: quadratic formula). One of these points will be $2$ (the other is irrelevant; can you figure out why?).

You do not need the squeeze theorem here.

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If there is a limit $x \ne 0$ then $x=3-2/x$, giving $x^2=3x-2$, giving $x=1$ or $x=2$. Since $a_1=3/2$ and the sequence is increasing ,we have $3/2 \le a_n<3$ by induction on $ n$. Hence the limit $x$ exists (because $3$ is an upper bound for the increasing sequence) and $ x>a_1>1$, so $x= 2$.

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