Proving that a sequence converge to $\sqrt{2}$

Question

I came upon this question while working out.

A sequence $u_n$ is defined as :

$u_1 = 1$ and $u_{n+1} = \frac12(u_{n} + 2/u_{n})$.

We have to show that this converge to square root of 2

I started like this to get a telescopic sequence

$ u_{n+1} = [(u_{n} - \sqrt{2})^2 + 2\sqrt{2}u_{n}]/2u_{n}$

from that I have got,

$ u_{n+1} = \sqrt{2} + [(u_n - \sqrt{2})^2]/2u_{n} $

I can't figure out how to go forward , I think my approach is wrong , can anyone please explain a method to get the answer?

Thank you so much !

And https://math.stackexchange.com/questions/338098/convergence-of-the-sequence-x-k1-frac12x-k-fracax-k — Jean-Claude Arbaut, Jun 06 '19 at 06:15

Michael Rozenberg · Accepted Answer · 2019-06-06T06:09:13.977

2

By induction easy to see that $u_n>0$ and by AM-GM for all $n\geq1$ we obtain: $$u_{n+1}=\frac{1}{2}\left(u_n+\frac{2}{u_n}\right)\geq\frac{1}{2}\cdot2\sqrt{u_n\cdot\frac{2}{u_n}}=\sqrt2.$$ Also, $$u_{n+1}-u_n=\frac{2-u_n^2}{2}\leq0.$$ Can you end it now?

edited Jun 06 '19 at 06:09

answered Jun 06 '19 at 05:51

Michael Rozenberg

194,933

You will need the fact that $$u_n\geq 0$$ for all $n$. – Dr. Sonnhard Graubner Jun 06 '19 at 06:07
@Dr. Sonnhard Graubner It's obvious by induction. – Michael Rozenberg Jun 06 '19 at 06:08
@Michael Rozenberg Just a little curious: What version of the AM-GM inequality do you use? Why not directly $(u_n+2/u_n)/2\ge\sqrt{u_n\cdot2/u_n}$? – Chrystomath Jun 06 '19 at 07:01
@Chrystomath I like $x_1+x_2+...+x_n\geq n\sqrt[n]{x_1x_2...x_n},$ where $x_i\geq0.$ – Michael Rozenberg Jun 06 '19 at 08:14
ok sir, so I can proceed with the Cauchy sequence theory now , is not it ? @MichaelRozenberg – Lasan Manujitha Ukwaththa Liya Jun 06 '19 at 11:38
No! We proved that our sequence decreases and bounded. Now, we can use Weierstrass – Michael Rozenberg Jun 06 '19 at 12:30
can you please explain sir , I'm still stuck with that , so many people have answered it in so many ways @MichaelRozenberg – Lasan Manujitha Ukwaththa Liya Jun 06 '19 at 14:52

Proving that a sequence converge to $\sqrt{2}$

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