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I have a sequence defined by $a_1=1$ and $a_{n+1}=3-\frac{1}{a_n}$. I need to use induction to show that the sequence is increasing and $a_n<3$ for all $n$.

Also to deduce that $a_n$ is convergence and find its limit.

So far I have found: $a_n<a_{n+1}<3$ and $a_{n+1}>0$

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

But not really sure how to go any further even if this is the right way to go in the first place? Any input would be greatly be appreciated!

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You have already done most of the work. By showing that $1 \leq a_n < a_{n+1} < 3$ where $n \in \mathbb{Z}_+$, you have demonstrated that the sequence is increasing and bounded above by $3$. Now the Monotone Convergence Theorem says that any increasing sequence that is bounded above converges; indeed it converges to the supremum of $\{a_n\}_{n \in \mathbb{Z}_+}$. So you now know that $\lim\limits_{n \to \infty} a_n = \alpha \in \mathbb{R}$ for some $$\boxed{1 \leq \alpha \leq 3}$$

To find $\alpha$, note that $f(x) = 3 - \frac{1}{x}$ is a continuous real valued function for real $x > 0$. So, by continuity, and the fact that $\alpha > 0$, we can put limits inside the function like so: $$\lim\limits_{n \to \infty}f(a_n) = f(\lim\limits_{n \to \infty}a_n)$$ But the LHS is just $\lim\limits_{n \to \infty}f(a_n) = \lim\limits_{n \to \infty}(3 - \frac{1}{a_n}) = \lim\limits_{n \to \infty} a_{n + 1} = \alpha$ while the RHS is just $f(\lim\limits_{n \to \infty}a_n) = f(\alpha) = 3 - \frac{1}{\alpha}$. So, putting LHS $=$ RHS we get $$\alpha = 3 - \frac{1}{\alpha}$$ $$\boxed{\alpha^2 - 3 \alpha + 1 = 0}$$ Now, solve this quadratic equation to get the value of $\alpha$. You will get two values but only one of them is the answer. See if you can eliminate the wrong one using the properties of $\alpha$.

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  • $\begingroup$ Thank you very much! $\endgroup$ – Olly Reynolds Apr 20 at 11:54
  • $\begingroup$ No problem, you're welcome. $\endgroup$ – ZeroXLR Apr 20 at 11:56
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Let $P(n)$ be the assertion “$a_n\in\left[1,\frac{3+\sqrt5}2\right)$ and $a_{n+1}>a_n$”.

If $n=1$, then this is true, since $a_1=1$ and $a_2=2>1=a_1$.

Let $n\in\mathbb N$ and assume that $P(n)$ holds. Then:

  • $a_{n+1}=3-\frac1{a_n}<3$. since $a_n\geqslant1>0$;
  • $a_{n+1}\geqslant1$, since $a_{n}\geqslant1$ and so $3-\frac1{a_n}\geqslant3-\frac11=2$.
  • $a_{n+2}-a_{n+1}=3-\dfrac1{a_{n+1}}-a_{n+1}=\dfrac{3a_{n+1}-1-{a_{n+1}}^2}{{a_{n+1}}^2}=\dfrac{\varphi(a_{n+1})}{{a_{n+1}}^2}$, where $\varphi(x)=-x^2+3x-1$. But the roots of the quadratic polynomial $\varphi(x)$ are $\dfrac{3\pm\sqrt5}2$ and so, since $a_{n+1}$ is between them and the coefficient of $x^2$ in $\varphi(x)$ is negative, $\varphi(a_{n+1})>0$.

So, $P(n+1)$ is proved.

What I wrote above is an inductive proof of the fact that we always have $P(n)$. And it is very easy now to conclude that $\lim_{n\to\infty}a_n=\dfrac{3+\sqrt5}2$.

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  • $\begingroup$ Great explanation thank you $\endgroup$ – Olly Reynolds Apr 20 at 11:54

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