Bounding above $a_{n+1}=\sqrt{16+|a_n|},\ a_0=4$

Question

I'm trying to prove $a_n\leq{5}$ with $a_{n+1}=\sqrt{16+|a_n|},\ a_0=4$

With induction $a_0<5$, and taking $a_k\leq5$ as the assumption. Then $a_{k+1}$

$a_{k+1}=\sqrt{16+|a_k|}$

$$\begin{align}\sqrt{16+|a_k|}&\leq5\\16+|a_k|&\leq25\\|a_k|&\leq9\end{align}$$ But this doesn't match up with my assumption. Is it a coincidence that $a_0+5=9$ also?

Answer 1

Though your general idea is on the right track, you have assumed what you wanted to prove. As a base case, you do indeed have $a_0= 4 < 5$.

Now assume that $a_k < 5$ for some $k \geq 0$. You now need to show that $a_{k+1} < 5$. Your first line is $a_{k+1}= \sqrt{16 + |a_k|} < 5$, but this is what you wanted to show! By writing this, you have started with the desired result.

You need to start with what you know: $a_k < 5$ (by assumption). You want to show that $a_{k+1} < 5$. How do we get from $a_k < 5$ to $a_{k+1} < 5$? Well, how does one get from $a_k$ to $a_{k+1}$? We use the recurrence definition! So we have... $$ \begin{split} a_k &< 5 \\ |a_k|&< |5| \\ |a_k|&< 5 \\ 16 + |a_k|&< 16 + 5 \\ 16 + |a_k|&< 21 \\ \sqrt{16 + |a_k|} &< \sqrt{21} \\ \sqrt{16 + |a_k|} &< \sqrt{21} < \sqrt{25} \\ \sqrt{16 + |a_k|} &< \sqrt{21} < 5 \\ \sqrt{16 + |a_k|} &< 5 \\ a_{k+1}&< 5 \end{split} $$ Then the result follows by induction.

NOTE. There are two steps that (at this level) require justification. In the first step, I said $a_k < 5$ so $|a_k| < |5|= 5$. Why? Why can't we have $a_k= -100$ so that $a_k < 5$ but then $|a_k|= |-100|= 100 \not< 5$? There is something you need to claim/prove about the $a_k$ so that this step is justified.

The second is the step where we have $16 + |a_k|< 21$ and claim that $\sqrt{16 + |a_k|}< \sqrt{21}$ (though we used this also when we showed $\sqrt{21} < \sqrt{25}$). If $0 < a < b$, why is it the case that $0 < \sqrt{a} < \sqrt{b}$? This also requires justification. I leave these two small gaps to you.

Final Note. If you can show that the $a_k$ is an increasing sequence (or that it is increasing after some $k$), you then have that it is bounded and increasing, you can then conclude...

The sequence $\{a_k\}$ converges by the Monotone Convergence Theorem.

But then we know that $\lim_{k \to \infty} a_k$ exists. You can call this limit $L$. But then...

$a_{k+1}= \sqrt{16 + |a_k|}$ so that $\lim_{k \to \infty} a_{k+1}= \lim_{k \to \infty} \sqrt{16 + |a_k|}$. But because $\lim_{k \to \infty} a_{k+1}= L$ and $\lim_{k \to \infty} \sqrt{16 + |a_k|}= \sqrt{16 + |L|}$. Then we have $L= \sqrt{16 + |L|}= \sqrt{16 + L}$. This is an equation you can solve for $L$! There will be two solutions - only one of which is the limit. (Which?) This still requires two small justifications: why is $|L|= L$ and why $\lim_{k \to \infty} \sqrt{16 + |a_k|}= \sqrt{16 + |L|}$. But these should be routine to justify.

Answer 2

$a_n$ is strictly increasing so $a_m\le a_\infty$ for all m

Note: $a_n$ is always positive that mod is irrelevant

Taking $b=a_\infty$ and applying limit to infinity on both sides of the recurrence relation

$$b^2=16+b\\ \ \\ b=\frac{1+\sqrt{65}}{2}\le5$$

Note: In your method for induction you need to take $a_n\le5$ not $a_{n+1}$

So you get

$$ a_{n+1}=\sqrt{16+a_n}\le\sqrt{21}\le5 $$