1. ## Sequence question

Question:
"Show that the sequence defined by a_1 = 1 a_(n+1) = 3 - 1/(a_n)
is increasing and a_n < 3 for all n. Deduce that {a_n} is convergent and find its limit."

Answer (from the back of the book):
1/2 * (3 + sqrt(5)

I notice that it is increasing because as a_n becomes larger the fractional part approaches 0 and therefore 3 - 0 = 3 so the limit should be 3 but my answer does not agree with the back of the book. It also seems convergent to me because it is bounded between [2,3) and it is monotonic (increasing) because the larger the the number you put on the denominator of the fractional part, the less you are subtracting and therefore the larger the final number is.

If someone could explain what I am doing wrong, I would really appreciate it!

2. Originally Posted by s3a
Question:
"Show that the sequence defined by a_1 = 1 a_(n+1) = 3 - 1/(a_n)
is increasing and a_n < 3 for all n. Deduce that {a_n} is convergent and find its limit."
$\displaystyle a_1 = 1$

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

To show it is increasing use induction...

$\displaystyle a_2= 3 - \frac{1}{a_1} = 3 - 1 = 2.$

$\displaystyle a_1 < a_2$ there is your base case.

Now show $\displaystyle a_n \leq a_{n+1}$

Now to show it is bounded.

$\displaystyle a_n \leq a_{n+1} \Rightarrow a_n - (3 - \frac{1}{a_n}) \leq 0...$

Now, since it is increasing and bounded above, it has a limit. This is an important theorem in Real Analysis.

3. Use induction.
Suppose that $\displaystyle a_{N+1}\le a_N <3$ then $\displaystyle \frac{1}{a_{N+1}}\ge\frac{1}{a_N}>\frac{1}{3}$.
But that implies that $\displaystyle 3-\frac{1}{a_{N+1}}\le 3-\frac{1}{a_N}< 3-\frac{1}{3}$.