Two Questions to Sequence.

1. Mar 2, 2008

Ka Yan

1. Is it possible for any real sequence {Sn} such that Sn > 0, for all n, and that lim sup Sn = $$\infty$$, while its arithmetic means an, definded as an = (S0 + S1 + ... + Sn)/(n+1) , (n = 0, 1, ...), such that lim an = 0 ?

2. How can I prove that the Newton's recursion formular xn+1 = (xn + a/xn)/2 converges to $$\sqrt{a}$$, if chosen x1 > $$\sqrt{a}$$ ?

Thks.

Last edited: Mar 2, 2008
2. Mar 2, 2008

Dick

1. Sure. A sequence can have very large terms without having large mean, can't it? 2. Consider f(x)=(x+a/x)/2. Then f(x_n)=x_n+1. You might want to think about where f(x)-x is increasing or decreasing.

3. Mar 3, 2008

Ka Yan

For the 1st question, could you please offer a example?
And thanks for giving me a hint to Q2.

4. Mar 3, 2008

HallsofIvy

Staff Emeritus
?? Dick's hint was to question 1! (That's why he put the big "1" in front!)

As for 2, separate it into two subsequences, n odd and n even. You should be able to show that for n odd, {xn} is a decreasing sequence of numbers larger than $\sqrt{a}$ and so converges while, for n even, {xn} is an increasing sequence of numbers less than $\sqrt{a}$ and so converges. Then use the recursion equation to show that the limit of each is $\sqrt{a}$.

5. Mar 3, 2008

Ka Yan

Erm... Thank you Professor Halls, for the solving of Q2. And Dick's SECOND hint did for Q2, since he put a "2" before writing the second sentence.

Thank you. Both you two.

Last edited: Mar 3, 2008
6. Mar 3, 2008

Dick

Hi Halls. It looks to me like the sequence is just plain decreasing isn't it? And to Ka Yan, set a_1=1, a_10=2, a_100=3, a_1000=4 etc. (other terms zero). Can you show the sequence of means approaches 0?

7. Mar 3, 2008

Ka Yan

In fact, for Q1, as anothter example, I can just simply let Sn = ln(n+1), and there I will get it.

8. Mar 3, 2008

Dick

Nice try. But I don't think the mean of that Sn goes to zero. You can approximate the sum of the Sn by an integral over n. The integral is (n+1)ln(n+1)-n-1. Divide by n+1 and it still goes to infinity.

9. Mar 4, 2008

Ka Yan

But mister, I wonder if I can apply the mean by L'Hospital's Law.
I made up that example, because I worked it up with the L. Law, and found it goes to 0, but I didn't quite sure that if an can satisfy the condictions so that the law works.

10. Mar 4, 2008

Dick

l'Hopital's Law would tell you lim ln(n+1)/n goes to zero as n->infinity. But you want to show lim sum(ln(n+1))/n goes to infinity. Think about it. It's a mean, an average. No Sn that increases monotonically to infinity can work. It has to increase and decrease. BTW 'misters' aren't necessary in the forum.