Considering the succession


$ b_{n+1}= \frac{b_n}{1-b_n}$

Prove by induction that $b_n<0$ for any natural number $n\ge2$

What I've done is: Prove for $n=1$:

$$b_{1+1}=b_2=\frac{b_1}{1-b_1}=\frac{2}{1-2}=-2$$ Which by the assumption made in the exercise, it's correct. Now, to prove that $b_n<0$, I only substituted. Like this:

Knowing that $b_n$ is like this: $$b_{n}=\frac{b_{n-1}}{1-b_{n-1}}$$

For $n=2$: $$b_{2}=\frac{b_{2-1}}{1-b_{2-1}}=\frac{b_{1}}{1-b_{1}}=\frac{2}{1-2}=-2$$

Which confirms the theory exposed in the exercise above.

Just in case, I tried it with $n=3$ and the result still came up with a negative value. $$b_{3}=\frac{b_{3-1}}{1-b_{3-1}}=\frac{b_{2}}{1-b_{2}}=\frac{-2}{1+2}=\frac{-2}{3}$$

Is this enough to prove the exercise? Or is it all wrong? Thanks for the help!

The exercise also asks if the succession $b_n$ grows. What I done was subtract $b_{n+1}-b_{n}>0$ (bigger than zero so it grows) Is it ok?

  • 2
    $\begingroup$ You have shown that $b_2$, and $b_3$ are negative, but your task is to prove that for any natural number $n\ge2$ that $b_n$ is negative. If you did that, you would know, for example, that $b_{3728457}$ must be negative, but does your proof make you sure of that? $\endgroup$ – Steve Kass Jun 25 '18 at 21:41
  • $\begingroup$ Yes, I understand but I've been trying to prove this exercise for over an hour, could you help me to get there please? I would really appreciate it. Thanks $\endgroup$ – Agapita Jun 25 '18 at 21:52
  • $\begingroup$ So if I stand on a street corner and a man walks by and the person next o me says "Everybody who walks by now will be a woman". So I check and the first person is a woman. Then the next. For good measure I wait for a third. It's a woman. And your are asking is that enough for a proof? No. Of course it is not $\endgroup$ – fleablood Jun 25 '18 at 22:32

You need to prove the statment for every $n$, not just for $n=1$, $n=2$ and $n=3$. So the natural question is how to prove some statment for every natural number (greater than two) when there is infinetly many natural numbers? How to check the statment for infinetly many cases?

Solution is to use mathematical induction. Let us say that you have some statement that depends on $n$ (like one you have in your question). Induction demands $3$ steps.

Step #1: Base of induction. You need to check your statement for one natural number. So let us check your statement for $n=2$ (because you need to check it for every $n$ greater than two): $ b_{2}=\dfrac{b_{1}}{1-b_{1}}= \dfrac{2}{1-2}=-2<0.$

Step #2: Assumption of induction. In this step we make an assumption. We assume that your statement works for some arbitrary number $k$. That means we assume that for some arbitrary number $k$: $b_{k}<0$.

Step #3: Inductive step. Now, we proove that then it must work for $k+1$. So here it is: $ b_{k}<0 \implies \dfrac{b_{k}}{1-b_{k}} <0 $, this is because $b_{k}$ is negative. So $1-b_{k}$ will be positive so when you divide negative ($b_{k}$) and positive($1-b_{k}$) you get again negative (that means less than zero). But that means $b_{k+1}$ must be negative because: $0>\dfrac{b_{k}}{1-b_{k}}=b_{k+1}$. Now what we have proved is that if yor statement works for some arbitrary $k$ that it works for $k+1$. But your statement works for number chosen in base of induction, namely number 2. That means that it has to work for number $2+1=3$. But now when it works for number $3$ it must work for $3+1=4$ and so it will work for every natural number.

So I will now solve second part when you need to see that sequance gets bigger. Sequance gets bigger only after $n\geq 2$.

Step #1. $b_{2}=-2, b_{3}=-\dfrac{2}{3} \implies b_{3}>b_{2}.$

Step #2. We make assumption that for some arbitrary $k$ statement $b_{k+1}>b_{k}$ is true.

Step #3. Let us first note: $b_{k+1}=\dfrac{b_{k}}{1-b_{k}} \implies \dfrac{1}{b_{k+1}}=\dfrac{1-b_{k}}{b_{k}}=\dfrac{1}{b_{k}}-1.$ Now we have: $b_{k+1}>b_{k} \implies \dfrac{1}{b_{k+1}}<\dfrac{1}{b_{k}} \implies \dfrac{1}{b_{k+1}}-1<\dfrac{1}{b_{k}}-1 \implies \dfrac{1}{b_{k+2}}<\dfrac{1}{b_{k+1}} \implies b_{k+2}>b_{k+1}.$

And then by mathematical induction it is proved that for every $n>2$ sequence get bigger.

  • $\begingroup$ @RM97 OMG! How do you even come up with this?! Thank you so much. Nobody has ever explain me induction so well in my life! $\endgroup$ – Agapita Jun 25 '18 at 22:31
  • $\begingroup$ @dxiv thanks for suggestion, I did not explain that part well. I corrected my mistake and made more correct and more pedagogical explanation. $\endgroup$ – Thom Jun 26 '18 at 0:38
  • $\begingroup$ @RM97 That works (+1). To be complete, you could also mention that $b_{k+1} \gt b_{k} \implies \dfrac{1}{b_{k+1}} \lt \dfrac{1}{b_{k}}$ only holds if $\,b_k\,$ and $\,b_{k+1}\,$ have the same sign, which was proved in the first part. $\endgroup$ – dxiv Jun 26 '18 at 0:44

Hint: $\;\dfrac{1}{b_{n+1}} = \dfrac{1-b_n}{b_n} = \dfrac{1}{b_{n}} - 1\,$, so if $\,b_n\,$ is negative then $\,\ldots$

Also, $\;\dfrac{1}{b_{n+1}} = \dfrac{1}{b_{n}} - 1 \lt \dfrac{1}{b_n}\,$, so $\,\dfrac{1}{b_n}\,$ is decreasing, and therefore $\,b_n\,$ is $\;\ldots$

  • $\begingroup$ Thanks, it really helps a lot. I just have one question...Why do you put the succession in a fraction? $\endgroup$ – Agapita Jun 25 '18 at 22:22
  • 1
    $\begingroup$ @Agapita That's just a convenient way to "separate" the $\,b_{n+1}\,$ and $\,b_n\,$ terms, and "see" the relation between them. $\endgroup$ – dxiv Jun 25 '18 at 22:25

You assume that for one particular $n$ that you now that $b_n< 0$. So now you have to prove that $\frac {b_n}{1 - b_n} < 0$. Now $b_n < 0$ so that means $\frac {b_n}{1-b_n} < 0$ if and only if $1 - b_n > 0$. And that is true if and only if $b_n < 1$. Which... it is... because $b_n < 0$.

So the proof is:

Proposition: For any $n \ge 2$ $b_n < 0$.

Proof of $n = 2$. $b_2 = \frac {b_1}{1-b_1} = \frac 2{1-2} -2 < 0$ so it is true for $n= 2.

If it is true for $n=k$ then here is the proof that it is true for $n=k+1$.

$b_k < 0$ so $1-b_k > 1 > 0$. And $b_k < 0$ so $b_{k+1} = \frac {b_k}{1-b_k} < 0$.

That's it. Intitial step. Proposition is true for $n=2$. Inductive step. If true for any $n = k$ it will be true of $n =k+1$. So it is true for all $n \ge 2$.


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