Finding Bounds, Proof by Induction

Question

I am trying to prove the upper bound is at most 12 and the lower bound is at least 1/2 for a sequence where $a_1$ = $1$ and $a_{n+1} = 5 + 2/a_n$, using induction. I tried it so my base case is $a_1 = 1$, and I established that $a_k > 1/2$. From here, $1/a_k < 2$ and $2/a_k < 4$. However, this is going nowhere and I'm wondering if this is even the correct solution.

Answer 1

We shall prove that $5<a_n<6$ for all $n\in \{3,4,5,\cdots\}$. We have that

$$a_2=5+\frac{2}{a_1}=5+2=7$$

$$a_3=5+\frac{2}{7}=5.28571$$

So the proposition holds for $n=3$. Now suppose it hold for some $n\geq 3$ and consider $a_{n+1}$: Obviously

$$a_{n+1}=5+\frac{2}{a_n}>5$$

However, we also know that

$$a_{n+1}=5+\frac{2}{a_n}<5+\frac{2}{5}=5.4<6$$

Thus, $5<a_n<6$ for all $n\in\{3,4,5,\cdots\}$. Additionally, $\frac{1}{2}<a_1,a_2<12$ which satisfy your bounds.

Answer 2

Hint: See that the sequence oscillates back-and-forth & converges to a particular value. This value is nothing but the continued fraction; (which could be solved by $5+2/x=x$) $$5+\frac{2}{5+\frac{2}{5+\frac{2}{...}}}=\frac{5+\sqrt{33}}{2}\simeq 5.3722813232$$

By this, you can see that the sequence hits a maximum & minimum possible values, in the process of 'oscillation', at $n=1$ & $n=2$. Or; $$a_1>a_n>a_2 \tag{for $n∈ℕ,n>2$}$$ $$7>a_n>\frac{37}{7}$$