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the question So I got this question and I was able to do up to (III) (a) however part b has been pretty difficult for me to complete and I can't seem to get it right . Any help ?

I tried considering $a_n - a_{n+1}$ and I got the required proof but I don't feel as though this method is logical enough

$a_n - a_{n+1} < 0$

$a_n < a_{n+1}$

TheSimpliFire
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user122343
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4 Answers4

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Guide:

  • Prove that $a_n>0$ if $a_1>0$.
  • Compute $a_{n+1}-a_n$ in terms of $a_n$, your goal is to check that the expression is positive.
  • check that the denominator is positive.
  • Show that the numerator is positive using the property that $0<a_n<2$.
Siong Thye Goh
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  • Thanks that makes sense to me and I was able to get through the proof. I have one question , how would you write a proof that $a_n$ > 0 ? I can clearly see it's possible but I would like to know a formal way of writing this – user122343 Aug 13 '18 at 04:20
  • @user122343 You have that $(a_n)$ is a sequence of positive numbers as an assumtion. There is no proof required. :) – Cornman Aug 13 '18 at 04:21
  • you are given that, but if you want, one possible way is to use mathematical induction. – Siong Thye Goh Aug 13 '18 at 04:22
  • @cornman thanks so much !!!! I totally glanced over that and it made the proof so much easier – user122343 Aug 13 '18 at 04:22
  • @cornman do you know where I can find inequality proof questions similar to these ? – user122343 Aug 13 '18 at 04:23
  • @user122343 I can check my own homeworks for similar questions, but they might be harder. Are you by any means from germany? – Cornman Aug 13 '18 at 04:27
  • @cornman unfortunately not . I appreciate the gesture. – user122343 Aug 13 '18 at 04:28
  • @user122343 Maybe not exactly what you are searching, but there is this german pdf with many questions, which are asked to be proved by induction. On page 4 under "E) (rekursive) Folgen" you maybe find something appropriate. http://www.emath.de/Referate/induktion-aufgaben-loesungen.pdf The questions 1) to 4) should be easy to understand without knowing german. Thre phrase "dann gilt" translates to "then holds". Therefor the term at the end is to be proven. I also checked old homeworks of mine, but did not find fitting questions. – Cornman Aug 13 '18 at 04:35
  • @cornman thanks so much this will definitely help ! – user122343 Aug 13 '18 at 04:39
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Alt. hint: $\;a_{n+1}=\dfrac{4(a_n+1\color{red}{+3-3})}{a_n+4}=4 - \dfrac{12}{a_n+4}\,$, and therefore:

$$\require{cancel} a_{n+1}-a_{n}=\left(\cancel{4} - \dfrac{12}{a_n+4}\right) - \left(\cancel{4} - \dfrac{12}{a_{n-1}+4}\right) = \dfrac{12(a_n-a_{n-1})}{(a_{n}+4)(a_{n-1}+4)} $$

It follows that $\,a_{n+1}-a_n\,$ and $\,a_n-a_{n-1}\,$ have the same sign, so the entire sequence is monotonic, and the direction of monotonicity is given by the sign of $\,a_2-a_1\,$, in this case positive so the sequence is increasing.

dxiv
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Item III.a. we use II. result $a_{n+1} - 2 = 2 - \frac{12}{4 + a_n}$, and $a_n < 2 \iff a_n + 4 < 6$, then $\frac{12}{4 + a_n} > 2$ Therefore $a_{n+1} - 2 = 2 - \frac{12}{4 + a_n} < 0 \iff a_{n+1} < 2$.

GinoCHJ
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So you can check your solutions.

We get $a_2=\frac{20}{11}$ and $a_3=\frac{31}{16}$.

For II):

It is $a_{n+1}-2=\frac{4(1+a_n)}{4+a_n}-2=\frac{4+4a_n-8-2a_n}{4+a_n}=\frac{2a_n-4}{4+a_n}=\frac{2(a_n-2)}{4+a_n}$

For III):

It is $a_{n+1}=\frac{4(1+a_n)}{4+a_n}\stackrel{a_n<2}{<}\frac{4(1+2)}{4+a_n}=\frac{12}{4+a_n}\stackrel{a_n<2}{<}\frac{12}{4+2}=2$. Hence $a_{n+1}<2$.

Finally:

We show, that $a_{n+1}-a_n>0$

We have $\frac{4(1+a_n)}{1+a_n}-a_n=\frac{4(1+a_n)-a_n(1+a_n)}{1+a_n}=\frac{4-a_n^2}{4+a_n}=\frac{(2+a_n)(2-a_n)}{1+a_n}>0$

Because $2+a_n>0$ and $1+a_n>0$ since $a_n>0$ (sequence of positiv numbers) Also $2-a_n>0\Leftrightarrow 2>a_n$ which is true.

Every factor greater than zero means the product is greater than zero.

Cornman
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