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Given is a sequence $a_n$. $a_1=7$ and $a_{n+1}=\frac{a_n^2+3}{2(a_n-1)}$ $n\in \Bbb N$. Prove that $a_n>a_{n+1}>3$ for all $n\geq 1$.

So I've proven that $a_n$ and$a_{n+1}$ are bigger than $3$ for all $n\geq 1$ by induction, but I'm stuck at proving that $a_n>a_{n+1}$.

Also, since this sequence is bounded below and monotone decreasing, I know it converges. But I can't just say that it converges to $3$ now, can I?

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    To answer the last question, no. Because you also have $a_n\gt 2\gt1.4\gt-1...$ – Shuri2060 Jul 14 '17 at 11:33
  • @Shuri2060 But I can say it's convergent right? –  Jul 14 '17 at 12:04
  • Not necessarily - depends on your reasoning. If you're just going by $(\forall n),,,a_n>3$ then you could have as your sequence $5,4,5,4,...$. However, if you're using $(\forall n),,,a_n>a_{n+1}>3$, then yes. – Shuri2060 Jul 14 '17 at 12:07
  • @Shuri2060 But we also know it decreases, since $a_n>a_{n+1}$ –  Jul 14 '17 at 12:08
  • The justification there is that the sequence is decreasing and bounded below. https://en.wikipedia.org/wiki/Monotone_convergence_theorem – Shuri2060 Jul 14 '17 at 12:08
  • Why wouldn't $a_n \cdot a_1 = 7$ imply that $a_n$ is a constant sequence? – user193319 Jul 14 '17 at 13:19
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    @user193319 It's a full stop after $a_n$. So We only know $a_1=7$ –  Jul 14 '17 at 13:53
  • I have found the limit $a_n\to\dfrac{1}{2} \left(1+\sqrt{13}\right)$ – Raffaele Jul 14 '17 at 18:57
  • @Raffaele But that's not possible since $a_n>3$ for every $n$ –  Jul 15 '17 at 12:02

3 Answers3

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we have $$a_n-a_{n+1}=a_n-\frac{a_n^2+3}{2(a_n-1)}=\frac{2a_n^2-2a_n-a_n^2-3}{2(a_n-1)}=\frac{(a_n+1)(a_n-3)}{2(a_n-1)}>0$$ since $$a_n>3$$

Dr. Sonnhard Graubner
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Given that you have prooved that $$a_n>3$$ for every $n=1,2,\dots$ we have, $$a_n>a_{n+1}\Leftrightarrow a_n>\frac{a_n^2+3}{2(a_n-1)}\overset{a_n-1>0}{\Leftrightarrow}2a_n(a_n-1)>a_n^2+3\Leftrightarrow a_n^2-2a_n-3>0$$ Now, let us consider the polynomial: $$p(x)=x^2-2x-3$$ We have: $$\Delta=(-2)^2-4(-3)=16=4^2$$ so, its roots are: $$x_{1,2}\frac{2\pm4}{2}=-1\mbox{ or }3$$ Since the leading coefficient is positive, we have that: $$p(x)>0\mbox{ for every }x>3$$ So, since $a_n>3$, we have that: $$p(a_n)>0\Leftrightarrow a_n^2-2a_n-3>0$$ which is equivalent to the wanted inequality.

Vassilis Markos
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$$a_{n+1}-3=\frac{a_n^2+3}{2(a_n-1)}-3=\frac{(a_n-3)^2}{2(a_n-1)}$$ and since $a_1>3$ by induction we obtain $a_n>3$ for all $n$.

In another hand, $$a_{n+1}-a_n=\frac{a_n^2+3}{2(a_n-1)}-a_n=\frac{(3-a_n)(a_n+1)}{2(a_n-1)}<0,$$ which says $a_n>a_{n+1}>3$ and we are done!

Michael Rozenberg
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