Recursive sequence problem: proofs by mathematical induction

[DivGradCurl]

Guys,

I'm trying to prove by induction that the sequence given by $$a_{n+1}=3-\frac{1}{a_n} \qquad a_1=1$$ is increasing and $$a_n < 3 \qquad \forall n .$$

Is the following correct? Thank you.

Task #1.
$$n = 1 \Longrightarrow a_2=2>a_1$$ is true.
We assume $$n = k$$ is true. Then,
$$3-\frac{1}{a_{k+1}} > 3-\frac{1}{a_k}$$
$$a_{k+2} > a_{k+1}$$ is true for $$n=k+1$$.
This shows, by mathematical induction, that
$$a_{n+1} > a_{n} \qquad \forall n .$$

Task #2
We already know that
$$a_1 < 3$$ is true.
We assume $$n=k$$ is true. Then,
$$a_k < 3$$
$$\frac{1}{a_k} > \frac{1}{3}$$
$$-\frac{1}{a_k} < -\frac{1}{3}$$
$$3-\frac{1}{a_k} < 3-\frac{1}{3}$$
$$a_{k+1} < \frac{8}{3} < 3$$
$$a_{k+1} < 3$$ is true for $$n = k+1$$. Thus,
$$a_{n} < 3 \qquad \forall n .$$

 

[fresh_42]

This is not written very well and some indices are wrong.
For the first step with $n=1$ we have $3 > 2 = a_2 = 3- \frac{1}{a_1} > 1 = a_1 > 0$.
Let us next assume we have shown $3 > a_n > a_{n-1} > 0$.

[Here we have to write the calculation 'backwards'.]
$$
3 > 3- \frac{1}{a_n} = a_{n+1} \stackrel{(1)}{>} 3 - \frac{1}{a_{n-1}} = a_n > 0
$$
where $(1)$ follows from: $a_n > a_{n-1} \Longrightarrow \frac{1}{a_{n-1}} > \frac{1}{a_n} \Longrightarrow -\frac{1}{a_n} > -\frac{1}{a_{n-1}}$

 

[Mark44]

In addition to fresh_42's comments, this line:

We assume $n=k$ is true. Then,
$a_k < 3$

You don't "assume" that n = k is true -- you assume that the proposition $a_n < 3$ is true for n = k.