Find a sequence convering to sqrt(2)

1. Dec 15, 2008

kidsmoker

1. The problem statement, all variables and given/known data

Find a sequence whose limit is $$\sqrt{2}$$.

2. Relevant equations

The work preceeding this was about using recurrence relations to find sequences with desired limits, so that's the method they want me to use.

3. The attempt at a solution

We can find the limit of the sequence given by

$$a_{n+1} = \frac{1}{2+a_{n}}$$ by noting that $$a_{n+1}$$ and $$a_{n}$$ both have the same limit. So we can write

$$l = \frac{1}{2+l}$$ and find the positive root of that: $$l = -1 + \sqrt{2}$$. This is the limit of the sequence.

I can then just add on one to each term to give a sequence with limit $$\sqrt{2}$$. Is this all they want me to do you think? Or is there a way to get write another sequence involving $$a_{n+1}$$ and $$a_{n}$$ which gives the desired answer?

Thanks.

2. Dec 15, 2008

Hurkyl

Staff Emeritus
Re: Sequences

There are gajillions of ways to answer this problem; yours seems fine.

I do have one question, though: were you merely asked to find a sequence, or are you also expected to prove that the sequence converges to $\sqrt{2}$? If the latter, I would like to point out that your work shows
If the sequence converges, then it converges to $\sqrt{2}$.​
However, you never showed that the sequence converges, and therefore you cannot conclude that it converges to $\sqrt{2}$.

3. Dec 15, 2008

kidsmoker

Re: Sequences

Yeah I see what you mean. If it was Analysis then i'd have to prove it by showing it's bounded and increasing or whatever. But this module doesn't seem to bother with that. Thanks.