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Here is a problem I've been working on. I am stuck and wondered if you guys could shed any light.

Let $a>0$ and $u_{1}>a$.

Consider the sequence $(u_{n})_{n=1}^{\infty }$ defined by:

$$ u_{n+1}\: :=\: \frac{1}{2}(u_{n}\:+\:\frac{a^2}{u_{n}})\;\;\;\;\forall n\in \mathbb{N} $$

Show that the sequence is convergent and that:

$$ \lim_{n \to \infty}u_{n}=a $$

Thank you in advance guys,

Deborah

Martin Sleziak
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Perat1980
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2 Answers2

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If $u_{n}$ converges, then $u_{∞+1} = u_{∞}$, so the simplest test is to set $u_{n+1} = u_{n}$ and then see if you can solve algebraically.

Simpson17866
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$$u_2:=\frac12\left(u_1+\frac{a^2}{u_1}\right)\le u_1\iff u_1^2\ge a^2\iff u_1\ge a\;\;\color{green}\checkmark$$

Inductively:

$$u_{n+1}=\frac12\left(u_n+\frac{a^2}{u_n}\right)\le u_n\iff u_n\ge a$$

But the last inequality can also be proved inductively:

$$u_{n+1}:=\frac12\left(u_n+\frac{a^2}{u_n}\right)\ge a\iff u_n^2+a^2\ge au_n\iff$$

$$u_n^2-au_n+a^2\ge 0\;\;\text{, and this is true iff the quadratic's discriminant is non-positive:}$$

$$\Delta:=a^2-4a^2=-3a^2<0$$

Thus, the sequence $\;\{u_n\}\;$ is monotone non-increasing and bounded below by $\;a\;$ so its limit exists, call it $\;U\;$ (0bserve that $\;U\ge a>0\;$ and, in particular, $\;U\neq 0\;$ ), and now a little arithmetic of limits:

$$U\leftarrow u_{n+1}:=\frac12\left(u_n+\frac{a^2}{u_n}\right)\rightarrow \frac12\left(U+\frac{a^2}{U}\right)\implies \ldots$$

DonAntonio
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  • Thanks for your help but, I don't see how you proved that $$u_{n} \geq a$$ – Perat1980 Nov 07 '13 at 16:12
  • Did you follow the lines after "Inductively" and until the expression for the discriminant $;\Delta;$? What isn't clear? – DonAntonio Nov 07 '13 at 17:28
  • I followed the first line up to the green tick. However, in the next line you assume that:$$u_{n} \geq u_{n+1}$$ – Perat1980 Nov 07 '13 at 17:39
  • No, @Perat1980: what I do that is $;u_n\ge u_{n+1}\iff\ldots;$ , see? Follow that, and you'll se this inequality is true iff the discriminant is negative, and since it is trivial it is then we're done! – DonAntonio Nov 07 '13 at 18:49
  • Ok I see, I agree with that now but surely to prove that $;u_n\ge u_{n+1};$ is true, since it is true iff $u_{n} \geq a$ you must prove that $u_{n} \geq a$ is true? Sorry if I'm being dumb. – Perat1980 Nov 07 '13 at 19:48
  • But that is what it's proved there! Observe the end of the line immediately after "Inductively"... – DonAntonio Nov 07 '13 at 19:58