We are given a sequence

$a_1=1,\ a_{n+1}=\frac{a_n^2+c}{2a_n}$ for all $n\geq 1\ (c>0)$

and we need to prove it's defined for all $n$ and converges (and to what value, but that's easy).

My attempt: I want to show that $a_n$ is bounded and thus is defined. I showed that for all $n\geq 2,\ a_n\geq \sqrt c$ (the limit of the sequence) so $a_n\geq \text{min}(1,\sqrt c)$, and I'm pretty sure $a_2=\frac{1+c}{2}$ is an upper bound, but I cannot show it really is/find another easier to prove upper bound.

I feel like I'm the the wrong direction, any help is welcomed!

  • $\begingroup$ Can you prove that this sequence decreases for $n\ge 2$? $\endgroup$ – Guy Oct 11 '16 at 12:44
  • $\begingroup$ Is it possible that you mean "sequence" rather than "series"? $\endgroup$ – Evan Aad Oct 11 '16 at 12:44
  • $\begingroup$ I think you must prove $\;a_n\neq0\;$ for all indexes in order to prove it is defined. Boundedness and etc. will come later. Also, I guess $\;c>0\;$ ...? $\endgroup$ – DonAntonio Oct 11 '16 at 12:44
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    $\begingroup$ For what it's worth, it seems that $$\forall n\in \mathbb N\left(a_n=\sqrt c\,\text{coth}\left(2^{n-1}\text{arctanh}(\sqrt c)\right)\right)_.$$ $\endgroup$ – Git Gud Oct 11 '16 at 13:05
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    $\begingroup$ It is Newton's method applied to $f(x)=x^2-c$. We have $$ x-\frac{f(x)}{f'(x)}=\frac{x^2+c}{2x}$$ and since $f(x)$ is a convex function, $a_n\to \sqrt{c}$ pretty fast. $\endgroup$ – Jack D'Aurizio Oct 11 '16 at 13:34

Let $f(x)=\frac{1}{2}\left(x+\frac{c}{x}\right)$ (I guess that $c>0$).

i) Note that if $x>0$ the $f(x)>0$, which means that $a_n>0$ (the sequence is well-defined).

ii) For $x>0$, $f(x)\leq x$ iff $x\geq \sqrt{c}$ (why), which means that if $a_n\geq \sqrt{c}$ then $a_{n+1}=f(a_n)\leq a_n$.

iii) If $x\geq \sqrt{c}$ then $f(x)\geq \sqrt{c}$ (why?) which means that if $a_n\geq \sqrt{c}$ then $a_{n+1}=f(a_n)\geq \sqrt{c}$.

Now $a_2=\frac{1+c}{2}\geq \sqrt{c}$, therefore, by ii) and iii), for $n\geq 2$, $a_n\in [\sqrt{c},a_2]$ and $a_n$ is decreasing.

  • $\begingroup$ I can prove that it's decreasing from $n=2$ and on, but I don't see how does this prove it's defined? (We assume this in the proof aswell) $\endgroup$ – Theorem Oct 11 '16 at 12:55
  • $\begingroup$ @Theoniix See my edited answer. $\endgroup$ – Robert Z Oct 11 '16 at 13:04
  • $\begingroup$ @Theoniix $a_n\not=0$ because if $x>0$ then $f(x)>0$ (if $c>0$), $\endgroup$ – Robert Z Oct 11 '16 at 13:07
  • $\begingroup$ So I just use induction to show it's well defined (which I can do in my own way in which I proved the inequallity)? $\endgroup$ – Theorem Oct 11 '16 at 13:10
  • $\begingroup$ @Theoniix Yes. Any more doubts? $\endgroup$ – Robert Z Oct 11 '16 at 13:23

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