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Given the recurrence relation $b_{1}=0$ and $$3b_{n+1} = \frac{b_{n}}{12} + \sqrt{\frac{17+b_{n}^{2}}{12}}$$

Show that this recurrence relation is increasing.

Note $36b_{n+1} = b_{n} + \sqrt{204+12b_{n}^{2}}$

So I have started as such:

I have found and proven $0\le b_{n}<1, \quad \forall n\ge1$. Thus

\begin{align} 36b_{n+1} - 36b_{n} &= \sqrt{204+12b_{n}^{2}} - 35b_{n} \\ &= \frac{204+12b_{n}^{2}-1225b_{n}^{2}}{\sqrt{204+12b_{n}^{2}} + 35b_{n}} \\ &= \frac{204-1213b_{n}^{2}}{\sqrt{204+12b_{n}^{2}} + 35b_{n}} \end{align}

But this is only increasing if $204-1213b_{n}^{2}>0 \iff b_{n}^{2} < 204/1213 \iff -\sqrt{204/1213}<b_{n}<\sqrt{204/1213}$. I know $0\le b_{n}<1$ but nothing more. What can I do?

user2850514
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1 Answers1

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Solve the inequality $f(x)=\frac{x}{36}+\sqrt{\frac{17+x^2}{108}}>x, x>0$, after a little bit rearrangement you get $x<\sqrt{\frac{204}{1213}}=c$, $c$ is also a solution of $f(x)=x$. It remains to prove every term is positive and less than $c$. Notice when $x>0$, $f(x)$ is strictly increasing so if $0<a<c$ we have $f(a)<f(c)=c$, by induction it's easy to show every term is less than $c$.