Prove the sequence has a limit value and find it's limit value

[Jarfi]

Homework Statement

Given the sequence: a(1) = 1, a(n+1)=0,5(a(n)+2/a(n))

n>=1

Homework Equations

I have found through speculations that the limit value is SQRT(2).

The Attempt at a Solution

I started by proving that for n>1; a(n+1) < a(n) and also proved that for n>1 the sequence is limited from below, by SQRT(2).

Now I have the conditions for the rule: if a function(or sequence(i think/hope)) f is decreasing, and is limited from below, it has the limit value;

Lim(f) = A when x-->infinity where A = inf(f(x)) where x is close to infinity.

Now the problem, I had two possible routes:

1: Find inf(f), I started by proving that for all n>1, a(n)>=SQRT(2)

then I made a K>SQRT(2) and tried proving that there existed an n+1>1 where a(n)<K, I am not sure of that but I think it can be done by choosing a domain for K and solving it for K, I got for a(n+1):

K-SQRT(k^2-2) < a(n) < K+SQRT(k^2-2) and this is where I was at so far.

2: Find the limit value of a(n) with conventional methods, whatever those are, because I have apparently done so a year ago when we were yet to explore sup and inf of functions, somehow you are suppost to get;

Lim(f) = SQRT(2) when x-->infinity but I'm not sure how you do that.







If anybody has a good tutorial or explanation on this, or help in general it would be greatly appreciated, I am suppost to give in my maths report tomorrow morning,

Thank you.

 

[Dick]

Homework Statement

Given the sequence: a(1) = 1, a(n+1)=0,5(a(n)+2/a(n))

n>=1

Homework Equations

I have found through speculations that the limit value is SQRT(2).

The Attempt at a Solution

I started by proving that for n>1; a(n+1) < a(n) and also proved that for n>1 the sequence is limited from below, by SQRT(2).

Now I have the conditions for the rule: if a function(or sequence(i think/hope)) f is decreasing, and is limited from below, it has the limit value;

Lim(f) = A when x-->infinity where A = inf(f(x)) where x is close to infinity.

Now the problem, I had two possible routes:

1: Find inf(f), I started by proving that for all n>1, a(n)>=SQRT(2)

then I made a K>SQRT(2) and tried proving that there existed an n+1>1 where a(n)<K, I am not sure of that but I think it can be done by choosing a domain for K and solving it for K, I got for a(n+1):

K-SQRT(k^2-2) < a(n) < K+SQRT(k^2-2) and this is where I was at so far.

2: Find the limit value of a(n) with conventional methods, whatever those are, because I have apparently done so a year ago when we were yet to explore sup and inf of functions, somehow you are suppost to get;

Lim(f) = SQRT(2) when x-->infinity but I'm not sure how you do that.

If anybody has a good tutorial or explanation on this, or help in general it would be greatly appreciated, I am suppost to give in my maths report tomorrow morning,

Thank you.

If you have proved that the sequence is decreasing and bounded below, then you know it has a limit L. Now finding what the limit is is easy. If you let n->infinity in a(n+1)=0.5(a(n)+2/a(n)) you get L=0.5(L+2/L). Just solve for L.

 

[Jarfi]

If you have proved that the sequence is decreasing and bounded below, then you know it has a limit L. Now finding what the limit is is easy. If you let n->infinity in a(n+1)=0.5(a(n)+2/a(n)) you get L=0.5(L+2/L). Just solve for L.

How do you know that when n -> infinity, a(n+1) = a(n) = L getting L=0,5(l+2/L)

Do I just say that infinity + 1 is equal to infinity?

 

[Dick]

How do you know that when n -> infinity, a(n+1) = a(n) = L getting L=0,5(l+2/L)

Do I just say that infinity + 1 is equal to infinity?

"infinity + 1" doesn't have much meaning. I definitely wouldn't say it that way. a(n) is the sequence {a(1),a(2),a(3),a(4),...}, a(n+1) is the sequence {a(2),a(3),a(4),...}. They aren't the same sequence but if the limit of a(n) is L, then the limit of a(n+1) is also L, isn't it? Just think about the definition of limit.

 

[Jarfi]

"infinity + 1" doesn't have much meaning. I definitely wouldn't say it that way. a(n) is the sequence {a(1),a(2),a(3),a(4),...}, a(n+1) is the sequence {a(2),a(3),a(4),...}. They aren't the same sequence but if the limit of a(n) is L, then the limit of a(n+1) is also L, isn't it? Just think about the definition of limit.

yeah, that explains it well. Can I word it like this; n+1=t-->infinity & n--> infinity, when n-->infinty. So that lim(a(t)) = L & lim(a(n)) = L when n->infinity.

Is this a "clean"/legit wording or should I try to assert what you said in a technical way?

 

[Dick]

yeah, that explains it well. Can I word it like this; n+1=t-->infinity & n--> infinity, when n-->infinty. So that lim(a(t)) = L & lim(a(n)) = L when n->infinity.

Is this a "clean"/legit wording or should I try to assert what you said in a technical way?

That's one way to say it. Substitute the variable t=n+1. I guess I wouldn't worry about stating it all that formally unless you really have to. It's sort of obvious. I'd just stay away from saying things like "infinity+1=infinity".