Limit of sequence

[Harmony]

1. The problem statement, all variables and given/known data
an+1=an/2 + 1/an

Prove that the above sequence converge and find the limit.
2. Relevant equations



3. The attempt at a solution
I have used Maple 12 to compute up to 10 term, using different initial value of a0. I found that the sequence is approaching square root of two when a0 is positive, and negative of square root of positive when a0 is negative. From the software, I know that the sequence is convergent, but it is rather difficult to find a mathematical proof.

Is it reasonable to say that an approximate to an+1 when n is approaching infinity? If so how can I prove it? Can I substitute both an and an+1 with ainfinity and find the limit?

Thanks in advanced.

[Pere Callahan]

If you know that your thus defined sequence (a_n)_{n\geq0} converges to some number c then the sequence (a_{n+1})_{n\geq0} converges to that same number.
Then you can use the continuity of the function x\mapsto \frac{x}{2}+\frac{1}{x} to take the limit of the equation a_n=\frac{a_n}{2}+\frac{1}{a_n} to obtain c=\frac{c}{2}+\frac{1}{c} which has the solution c=\sqrt{2} of course.

So if the sequence is convergent the limit is \sqrt{2}.

But in order to conclude that your sequence converges in the first place you need to work some more. How would you prove in general that a sequence is convergent?

[HallsofIvy]

1. The problem statement, all variables and given/known data
an+1=an/2 + 1/an

Prove that the above sequence converge and find the limit.
2. Relevant equations



3. The attempt at a solution
I have used Maple 12 to compute up to 10 term, using different initial value of a0. I found that the sequence is approaching square root of two when a0 is positive, and negative of square root of positive when a0 is negative. From the software, I know that the sequence is convergent, but it is rather difficult to find a mathematical proof.

Is it reasonable to say that an approximate to an+1 when n is approaching infinity? If so how can I prove it? Can I substitute both an and an+1 with ainfinity and find the limit?

Thanks in advanced.
Well, you shouldn't call it ainfinity. Rather, IF the sequence converges, say to some number A, then, taking limits on both sides of the equation A= A/2+ 1/A. From that, multiplying on both sides by A, A2+ A2/2+ 1 or (1/2)A2= 1 or A2= 2. Assuming that a0> 0 then A= \sqrt{2}.

That is, of course, IF the sequence converges. With that information, we can then use "monotone convergence" to show that it does, in fact, converge.

[statdad]

As a slight (I hope not too far-reaching) expansion on the previous correct ideas:

1. Try to show that the sequence is non-decreasing (i.e. a_{n+1} \ge a_n)
2. Show that the sequence is bounded above ( a_n \le M for some
positive number M )

These two steps together will be enough to prove that the sequence converges; you've already seen how to find the limit.