Let $(a_n)_{n \in \mathbb N}$ be a recursive sequence. It is defined as $a_1=1, \quad a_{n + 1} = \frac{4a_n}{3a_n+3}$.

I have to show by induction, that it is limited for $a_n \ge \frac{1}{3}$.

So as an induction step I have $n \to n + 1$. And the idea is to get $a_{n+2}$ and show somehow $a_{n+2} \ge \frac{1}{3}$

$$a_{n+1} = \frac{4a_n}{3a_n + 3} $$ $$a_{n+2} = \frac{4 \frac{4a_n}{3a_n + 3}}{3 \frac{4a_n}{3a_n + 3} + 3} = \frac{ \frac{16a_n}{3a_n+3} }{ \frac{12 a_n}{3a_n + 3} + 3}$$

Question: How should I go on or am I on the wrong path?

Edit: You can find other answers for the exact same question here.


You wonder if $a_{n+1}\geq {\frac{1}{3}}$, so if

$$\frac{4a_n}{3a_n+3}\geq {\frac{1}{3}}$$ but this is equivalent to $$ 4a_n \geq a_n+1$$ or $$3a_n\geq 1$$ which is true by induction assumption.

  • $\begingroup$ I think this only shows that if (for a given fixed $n$) your claim is true for $a_{n+1}$ then it is also true for $a_n$. This way you are moving backwards in the series and can prove your claim for only finite number of elements. But when using induction you want to move forwards, so eventually you can reach any element up to infinity $\endgroup$ – Andras Vanyolos Nov 11 '17 at 17:03
  • $\begingroup$ Of course, you have to read it from down to up. $\endgroup$ – Maria Mazur Nov 11 '17 at 17:04
  • $\begingroup$ Yes, indeed. It's an interesting trick I haven't seen before, works nicely for recursive series. $\endgroup$ – Andras Vanyolos Nov 11 '17 at 21:08

To show that $a_{n+2}\ge1/3$, given that $a_{n+1}\ge1/3$, I would just plot the formula $$ a_{n+2}=\frac{4a_{n+1}}{3a_{n+1}+3} $$ as a function of $a_{n+1}$. This is clearly a strictly increasing function for $a_{n+1}\ge1/3$, which has limit $4/3$, and it takes on its minimum value ($=1/3$) at $a_{n+1}=1/3$. So $a_{n+2}\in[1/3,4/3)$. So your claim holds and the induction step from $n+1$ to $n+2$ is proved.


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