Two Sequencing Questions for Converging Formulas

[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 formula xn+1 = (xn + a/xn)/2 converges to \sqrt{a}, if chosen x1 > \sqrt{a} ?

Thks.

 

[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.

 

[Ka Yan]

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

 

[HallsofIvy]

?? 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, {x_n} is a decreasing sequence of numbers larger than \sqrt{a} and so converges while, for n even, {x_n} 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}.

 

[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.

 

[Dick]

?? 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, {x_n} is a decreasing sequence of numbers larger than \sqrt{a} and so converges while, for n even, {x_n} 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}.

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?

 

[Ka Yan]

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

 

[Dick]

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

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.

 

[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.

 

[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.