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Consider the recursively defined sequence $a_n = 1$

$$a_{n+1} = 3 - \frac{1}{a_n} \text{ for n > 1}$$

Is the sequence convergent?

This is my attempt:

First, we prove that the sequence is positive, and monotonically increasing, using induction.

$\textbf{Base Case:}$ $a_1 = 1$ and $a_2 = 2$. $a_1 < a_2$ and $a_1,a_2 > 0$.

$\textbf{Inductive Hypothesis:}$ $a_n < a_{n+1}$ where $a_n, a_{n+1}$.

$\textbf{Inductive Step:}$ We prove that $a_n < a_{n+1} \Rightarrow a_{n+1} < a_{n+2}$.

By our induction hypothesis:

$$a_n < a_{n+1}$$

$$-a_n > -a_{n+1}$$

$$-\frac{1}{a_n} < -\frac{1}{a_{n+1}}$$ (True by our IH since $a_n, a_{n+1} > 0$ and thus, $-a_n, -a_{n+1}$ share the same sign).

$$3-\frac{1}{a_n} < 3-\frac{1}{a_{n+1}}$$

$$a_{n+1} < a_{n+2}$$.

Note that $a_{n+1} > 0$ by our IH, so $a_{n+2} > 0$.

Now we prove that the sequence is bounded.

Observe that $a_n$ is monotonically increasing, which means that $ - \frac{1}{a_n}$ is monotonically increasing as well and it is upped bounded by $0$. Thus, $3-\frac{1}{a_n}$ is upper bounded by $3$.

We have a sequence that is monotone and bounded. Hence, by the Monotone Convergence Theorem, This sequence converges.

I was wondering if this method is correct.

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  • $\begingroup$ FWIW, since neither of the answers actually says this: yes, your method is entirely correct (modulo the assertion that $a_n\gt 0$ for all $n$, which is straightforward to prove from what you've already got). $\endgroup$ – Steven Stadnicki Feb 7 '18 at 20:14
  • $\begingroup$ @StevenStadnicki that's not accurate, Michael points to the fact that $a_n>0$ is implied, but should be proved first. I presented a version where this is obvious. $\endgroup$ – rtybase Feb 10 '18 at 13:40
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A shorter approach, look at the function $f(x)=3-\frac{1}{x}$ because this function "generates" the sequence, i.e. $f(a_n)=a_{n+1}$. This function is ascending, because $f'(x)=\frac{1}{x^2}$ and $\frac{1}{3}<a_1<a_2 \Rightarrow 0< \color{red}{f(a_1)\leq f(a_2)} \Rightarrow \frac{1}{3}<a_1<\color{red}{a_2 \leq a_3}$ and by induction $\frac{1}{3}< a_{n} \leq a_{n+1}$. So, the sequence is ascending and positive.

Given $a_n > \frac{1}{3} > 0, \forall n$, then $\frac{1}{a_n}>0 \Rightarrow a_{n+1}=3-\frac{1}{a_n} < 3 \Rightarrow 0<a_{n+1} < 3, \forall n$. So the sequence in bounded.

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In your proof of $a_{n+1}>a_n$ you used that $a_n>0$, but it is not proven.

I like the following reasoning. $$a_{n+1}-\frac{3-\sqrt5}{2}=3-\frac{3-\sqrt5}{2}-\frac{1}{a_n}=\frac{3+\sqrt{5}}{2}-\frac{1}{a_n}=\frac{1}{\frac{3-\sqrt5}{2}}-\frac{1}{a_n}>0$$ by induction because $a_1=1>\frac{3-\sqrt5}{2}.$ $$a_{n+1}-\frac{3+\sqrt5}{2}=3-\frac{3+\sqrt5}{2}-\frac{1}{a_n}=\frac{3-\sqrt{5}}{2}-\frac{1}{a_n}=\frac{1}{\frac{3+\sqrt5}{2}}-\frac{1}{a_n}<0$$ by induction because $a_1=1<\frac{3+\sqrt5}{2}.$

Thus, for all natural $n$ we got: $$\frac{3-\sqrt5}{2}<a_n<\frac{3+\sqrt5}{2}.$$ In another hand, $$a_{n+1}-a_n=3-a_n-\frac{1}{a_n}=\frac{\left(a_n-\frac{3-\sqrt5}{2}\right)\left(\frac{3+\sqrt5}{2}-a_n\right)}{a_n}>0,$$ which says that $a$ converges.

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  • $\begingroup$ If $a_n=\frac32$, then it doesn't converge by your inequality. $\endgroup$ – Takahiro Waki Feb 10 '18 at 14:39
  • $\begingroup$ I proved that it converges to $\frac{3+\sqrt5}{2}.$ $\endgroup$ – Michael Rozenberg Feb 10 '18 at 15:55
  • $\begingroup$ If you say there is such a proof in this answer, you are not a mathematical graduate student. Also this question can be proved by even high school math. $\endgroup$ – Takahiro Waki Feb 10 '18 at 16:06
  • $\begingroup$ Yes, of course. My proof is very easy and it's a not hard problem. $\endgroup$ – Michael Rozenberg Feb 10 '18 at 16:11

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