# Limit of a sequence.

• Oct 18th 2008, 12:55 PM
Showcase_22
Limit of a sequence.
I wasn't sure which section this belonged to so I decided to post it here.

Quote:

Use a method to approximate $\sqrt 2$ using a sequence of rational numbers. Do the same for $\sqrt 3.$
I started it like this:

$a_1, \ a_2, \ a_3, \ ........... \ a_n \ ...... \ L$

$\frac{1}{a+a_1}+b, \ \frac{1}{a+a_2}+b \ ............\ \frac{1}{a+a_n}+b \ ..... \ \frac{1}{a+L}+b$

where L is the limit.

Both sequences must have the same limit:

$\frac{1}{a+L}+b=L$

$\frac{1+ba+bL}{a+L}=L$

$1+ba+bL=L^2+aL$

$L^2+aL-bL-1-ba=0$

$L^2+L(a-b)-1-ab=0$

Since we want L= $\sqrt 2$:

$f(\sqrt 2)=2+ \sqrt {2} (a-b)-1-ab=0$

Since we want this to be 0 a=b since the middle term cannot exist.

$1-b^2=0$

$b=1 \ or \ -1$

I want the positive $\sqrt 2$ so b=1.

Therefore the sequence becomes $a_{n+1}=\frac{1}{1+a_n}+1$. This sequence tends to $\sqrt 2$.

Using the same method to find one that tends to $\sqrt3$ gives:

$\frac{1}{a+L}+b=L$

$L^2+L(a-b)-1-ab=0$

Once again, the central term cannot be present at the end so a=b:

$f(\sqrt3)=3-1-b^2=0$

$b^2=2$

$b=\sqrt 2$

But b can only be a rational number so it doesn't make sense. I think i'm supposed to use the first answer to find the second one except the two sequences are entirely different!

So am I on the right lines?