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?

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

1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$
    – Monolite
    Commented Feb 10, 2016 at 22:21
  • 1
    $\begingroup$ 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. $\endgroup$
    – Jinmu You
    Commented Feb 10, 2016 at 22:40
  • $\begingroup$ 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? $\endgroup$
    – Monolite
    Commented Feb 11, 2016 at 10:24
  • $\begingroup$ 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 $\sqrt2$. $\endgroup$
    – Jinmu You
    Commented Feb 11, 2016 at 13:40

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