# Limit of sequence

1. Aug 27, 2008

### 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?

2. Aug 27, 2008

### 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?

3. Aug 27, 2008

### HallsofIvy

Staff Emeritus
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.

Last edited: Aug 27, 2008
4. Aug 27, 2008

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 $$M$$)