On a recursive sequence exercise $a_{n+2} = \frac{4 + 3a_n}{3 + 2a_n}$.

As part of an exercise I am given a sequence defined by $a_1 = 1$

and $$a_{n+1} = 1 + \frac{1}{1 + a_n}$$

I have noticed that the even sequence is decreasing and I want to prove this, the even sequence (and even the odd) will be given by

$$a_{n+2} = \frac{4 + 3a_n}{3 + 2a_n}$$

I proceed by induction and check the base case, then I suppose $a_{n+2} < a_{n}$ to prove that $a_{n+4} < a_{n+2}$. But through numerous substitutions I arrive nowhere.

I have the habit of posting my calculations but I lead myself only to dead ends, is induction the wrong way to go in this case? Could I bother you with a proof?

• something is missing. What is $a_2$ or $a_0$ for example ? Commented Feb 10, 2016 at 22:03
• @Svetoslav I see that it can be confusing, I will add part of the text. Commented Feb 10, 2016 at 22:05

It must go to dead ends, because the proposition is wrong.

Think about an inequality below : $$\frac{4+3x}{3+2x}<x$$ This is equivalent to $$(4-2x^2)(3+2x)<0$$ $$\therefore -1.5<x<-\sqrt2, x>\sqrt2$$

Let's pick $a_n=-1.45$ which is $-1.5<a_n<-\sqrt2$.

Then $a_{n+2}=-3.5$ and $a_{n+4}=1.625$.

So $a_{n+2}<a_n$ but $a_{n+4}>a_{n+2}$.

If you want even-decreasing sequence, you have to choose suitable initial value $a_n$ which makes all even terms $a_{2k}$ be larger than $\sqrt2$ so that they can satisfy the inequality above.

• Thank you for your answer, I did write the initial value even before my edit. Nonetheless could you explain to me how the inequality you wrote is related to my problem? Commented Feb 10, 2016 at 22:21
• Oh I didn't see the initial value. If $a_1=1$, the initial value will must be $a_2=1.5>\sqrt2$ and this value is good for your purpose. An appropriate $a_k$ that satisfies the inequality satisfies $a_{k+2}<a_k$, too. That is, $a_2=1.5$ for example, $a_4$ will be smaller than $a_2$. And it still larger than $\sqrt2$ (you can confirm by plotting $y=(4+3x)/(3+2x)$ and $y=x$) so $a_6<a_4, a_8<a_6, ....$. The left of the inequality is just a same form with your recursive equation! I just found out the condition when $a_{n+2}<a_n$ works. Commented Feb 10, 2016 at 22:40
• Thank you I now understand, but how do I know that my iterated $a_n$ will not go below $\sqrt{2}$ at a certain point? I am ok for the initial value but what about the rest? Commented Feb 11, 2016 at 10:24
• Consider graphs of $y=\frac{4+3x}{3+2x}$ and $y=x$ here. Since the blue graph is increasing at $x>\sqrt2$, when you pick any $a_k>\sqrt2$, always $a_{k+2}$ will be also larger than $\sqrt2$. Commented Feb 11, 2016 at 13:40