Expert: Sherry Wallin - 3/22/2011

Question
QUESTION: This was my reasoning using the comparison test:

e^(2*sqrt(k))> k^(3/2) since "exponential growth is faster than the power growth"
So (1/sqrt(k)) * e^(-2*sqrt(k))< 1/k^2 so our series converges by comparison with 1/k^2

Btw I have a final question if you can confirm my work. Thanks again for all the help!


Let x(n+1)= (3*x(n)) / (x(n))^2+2 n belongs in N, xo>=0 is some starting value

a) Show that if the sequence converges to a finite limit then this is 0 or 1.
b) Show that if 0<x(n)<1 then 0<x(n)<x(n+1)<1 and determine the limit behavior of the sequence when 0<x0<1.
c) Show that if 1<x(n)<2 then 1<x(n+1)<x(n) and determine the limit behavior of the sequence when 1<x0<2
d) What happens for x0>=2?


a) If x(n) tends to K in R then so does x(n+1) so K=3K/K^2+2 so K=0 or K=1 or K=-1 but we cross out K=-1 since x0>=0 and so ...>x2>x1>x0 (x0>0) or all x(n)=0 (x0=0)

b) x(n+1)-x(n)=(x(n)*(1-(x(n))^2)) / ((x(n))^2+2) which is >0 if 0<x(n)<1 and so x(n+1)>x(n). So the sequence is an increasing one with 0<x(n)<x(n+1)<1, since the only positive finite limit is 1.

When 0<x0<1 our sequence is increasing and so 1>...>x2>x1>x0>0. It is bounded above by 1 since is the only positive finite limit.

c) If 1<x(n)<2 then x(n+1)-x(n)<0 so x(n)>x(n+1) so our sequence is decreasing but >1 since is the only positive finite limit and so 1<x(n+1)<x(n)

When 1<x0<2 then 1<.....<x2<x1<x0 our sequence is decreasing and bounded below by 1.

d) If x0=2 then all x(n) are zero so we have a sequence full of zeros.
If xo>2 we observe that x1<x0 but 1>...>x3>x2>x1 and our sequence is bounded above by 1 since is the only possible finite limit.

ANSWER: Michael~

I like your comparison, very good. Just make sure you show that you multiplied both sides of the original inequality by sqrt k before moving onto (1/sqrt(k)) * e^(-2*sqrt(k))< 1/k^2.

In the other problems I am not sure about your notation. I think that you mean x_0 (x sub zero) by your notation but what I did in parenthesis is more appropriate or you could use x(0) like you do with x(n).In part a I don't know how you get k = 0, +-1 and this is why:
I don't think you can allow 0 because then 3k/k^2 = 3/k is undefined at 0. Second when I eliminate 0 as a possible then (and only then) can I multiply by k getting k^2-2k-3 = 0 which factors as (k+1)(k-3) = 0 which means that k = -1 0r 3 which is not what you want to show.

I can respond with more later, got to run to teach class now.

Math Prof

---------- FOLLOW-UP ----------

QUESTION: Ok, thanks for everything Dr. Wallin!

Answer
Michael~
For part b, rather than decipher what you wrote I will answer the question the way I would answer it and then put the burden on you to determine if you can follow my logic:

If 0<x_n<1, then 3x_n/x_n^2 = 3/x_n is certainly greater than 3,
thus x_(n+1) = 3x_n/x_n^2 + 2 > 3 + 2 = 5. This does not show that x_(n+1)< 1

Now let me comment on your way:
For starters the way you have the original problem written it says that x_(n+1) = [3x_n/x_n^2] now add 2, i.e., [3x_n/x_n^2] + 2. In your 'proof' below you are using x_(n+1) = [3x_n/(x_n^2+2)]. Which is it suppose to be? Again the way you wrote the original problem it should be the way I interpreted it, that is, do the division and then add 2. That may be why my 'proof' doesn't give the desired result so in that case you needed to write the problem so that the plus 2 was part of the denominator of the quotient. Now based on that assumption, I will comment on your 'proof':

'b) x(n+1)-x(n)=(x(n)*(1-(x(n))^2)) / ((x(n))^2+2) which is >0 if 0<x(n)<1 and so x(n+1)>x(n). So the sequence is an increasing one with 0<x(n)<x(n+1)<1, since the only positive finite limit is 1.

When 0<x0<1 our sequence is increasing and so 1>...>x2>x1>x0>0. It is bounded above by 1 since is the only positive finite limit.'

If you start with x(n+1)= (3*x(n)) / [(x(n))^2+2]  and subtract x_n from both sides you get: x(n+1)-x(n)=(x(n)*(1-(x(n))^2)) / ((x(n))^2+2) and while it is true that this quantity is greater than 0, which means the difference is positive and that x_(n+1) is larger than x_n, you still need to show that the numerator is always smaller than the denominator to show that x_(n+1)< 1

Other than the comments I have made this problem looks good.

Again I have to run, sorry!

Math Prof