I'm wondering how to explain something more formally about a bounded series.

Question

Here's the question:

A sequence $(a_n)$ is defined by $a_1 = 1$ together with the recursive formula $a_{n+1} = \frac{2+2a_n}{2+a_n}$

Prove, by induction or otherwise, that $1 ≤ a_n ≤ 2$ for all $n ∈ N$.

By using induction to get $a_{k+2}$ and setting $a_{k+2} >= a_{k+1}$, you find it is true for $\sqrt{2}$ but don't know the formal way of doing this. How can I then explain that the series stay between $[1,2]$?

Answer 1

Induction. Since: $$1 ≤ a_n ≤ 2 \implies 4 ≤ 2+2a_n ≤ 6 \;\;\;{\rm and}\;\;\;3 ≤ 2+a_n ≤ 4$$

so

$$3 ≤ 2+a_n ≤ 4\implies {1\over 4 } \leq {1\over 2+a_n}\leq {1\over 3}$$

so $$1={4\over 4}\leq {2+2a_n\over 2+a_n} \leq {6\over 3} =2$$

so $$1\leq a_{n+1} \leq 2$$

Answer 2

Expanding my comment: let us set $a_n = \frac{p_n}{q_n}$ and associate $a_n$ to the vector $v_n=(p_n,q_n)^T$.
Then the recurrence relation $$ a_{n+1} = \frac{Aa_n+B}{Ca_n+D} $$ can be written as $$ v_{n+1} = M v_n = M^n v_0,\qquad M=\begin{pmatrix}A & B \\ C & D\end{pmatrix}. $$ If we consider the Jordan normal form of $M$ and apply the Hamilton-Cayley theorem, we have that both the first and the second component of $v_n$ fulfill a linear recurrence relation, whose characteristic polynomial is the characteristic polynomial of the $M$ matrix. If $M$ has distinct eigenvalues $\lambda_1,\lambda_2$ it follows that $$ a_n = \frac{R\lambda_1^n+S\lambda_2^n}{T\lambda_1^n+U\lambda_2^n} $$ for some constants $R,S,T,U$ which can be found by interpolation.
In our case the eigenvalues of $\begin{pmatrix}2 & 2 \\ 1 & 2\end{pmatrix}$ are $2\pm\sqrt{2}$ and $$ a_n = \frac{\frac{1}{2\sqrt{2}}\left[(2+\sqrt{2})^n-(2-\sqrt{2})^n\right]}{\frac{1}{4}\left[(2+\sqrt{2})^n+(2-\sqrt{2})^n\right]}=\sqrt{2}\,\frac{1-(3-2\sqrt{2})^n}{1+(3-2\sqrt{2})^n}$$ is pretty obviously an element of $[1,2]$ which converges to $\sqrt{2}$ as $n\to +\infty$.
The continued fraction of $a_n$ is either $[1;2,2,\ldots,2,2]$ or $[1;2,2,\ldots,2,3]$.