I have been given the following problem: $$ a_1 = 1, a_{n+1} = 3-\frac{1}{a_n}\\ \text{ Find: }\lim\limits_{n \to \infty} a_n $$ I have already proved that the function is always increasing and bounded by 3. So I know the sequence will converge. However, I have not learned how to compute limits for successor functions, nor do I know how to transform this into a function of n.
My math book tells me the answer is $(3+\sqrt5)/2$, though I do not know how they reached this answer.
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$\begingroup$Now you know the limit exists, so you can assume it is $a$. Then take limit on both sides of $$ a_{n+1}=3-\frac{1}{a_n} $$ and you obtain: $$ a=3-\frac{1}{a} $$ Solve the equation, and you get two roots $\frac{3+\sqrt{5}}{2}$ and $\frac{3-\sqrt{5}}{2}$. You know the sequence is increasing and $a_1=1$ so $\frac{3+\sqrt{5}}{2}$ is the correct limit.
$\endgroup$ $\begingroup$Hint: in the limit as $n \rightarrow \infty$, $a_{n+1} \sim a_n$.
So set $a_{n+1}=a_n=a$. Then
$$a=3-\frac{1}{a} \implies a^2-3 a+1=0 \implies a=\frac{3\pm \sqrt{5}}{2}$$
Which root is correct? We want $a_n>1$ as befots the behavior of the sequence; the result follows.
$\endgroup$ $\begingroup$Let the limit be $a$. Then $$a=\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty} \left(3-\frac{1}{a_n}\right)=3-\frac{1}{a}.$$
After simplifying, we get the quadratic equation $a^2-3a+1=0$. Solve.
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