# Sequence Problem, Rudin's PMA Ch.3 #17

• Poopsilon
In summary, the problem is that for a given sequence r_n=x_n+2 - x_n, the sequence alternates sign, with it giving negative terms for n odd and positive terms for n even. My proof so far is that r_n=x_n+2 - x_n and r_n=x_n+1 - x_n and adding these two equations and setting r_n = x_n+2 - x_n produces: r_n = x_{n+2} - x_n = \frac{a - x_n^{2}}{1+x_n} + \frac{a - x_{n+1}^{2
Poopsilon
So for those of you who don't have the book the problem goes like this:

Fix a > 1, take x_1 > sqrt(a), and define:

$$x_{n+1}=\frac{a+x_n}{1+x_n}=x_n+\frac{a-x_n^{2}}{1+x_n}$$

a) Prove that x_1 > x_3 > x_5 > ...
b) Prove that x_2 < x_4 < x_6 < ...

Basically my strategy is to show that the sequence r_n=x_n+2 - x_n alternates sign, with it giving negative terms for n odd and positive terms for n even. My proof so far is as follows:

$$x_{n+1} - x_n = \frac{a - x_n^{2}}{1+x_n}$$ and $$x_{n+2} - x_{n+1} = \frac{a - x_{n+1}^{2}}{1+x_{n+1}}$$

adding these two equations and setting r_n = x_n+2 - x_n produces:

$$r_n = x_{n+2} - x_n = \frac{a - x_n^{2}}{1+x_n} + \frac{a - x_{n+1}^{2}}{1+x_{n+1}}$$

now substituting in $$\frac{a+x_n}{1+x_n}$$ for $$x_{n+1}$$ gives:

$$r_n = \frac{a - x_n^{2}}{1+x_n} + \frac{a - (\frac{a+x_n}{1+x_n})^2}{1+(\frac{a+x_n}{1+x_n})}$$

from here I simplify the right side until obtaining:

$$r_n = \frac{2x_n(a-x_n^{2})+2a}{(1+x_n)(2x_n+a+1)}$$

I have checked my work multiple times on this simplification and I cannot find any error. Yet if we set n=1 than clearly the denominator is positive, but if one chooses x_1 to be just a tiny bit greater than sqrt(a) than the numerator can be made positive as well, which contradicts the desired conclusion for part a).

I'm fairly certain my strategy is viable, hopefully someone can do the simplification and see if they are getting the same result. Its certainly possible that I'm missing something.

Last edited:
Hi Poopsilon!

I'm confused … first you put rn = xn+2 - xn,

then you put rn = xn+1 - xn

Use xn+2 - xn = (xn+2 - xn+1) plus (xn+1 - xn).

Sorry, you answered really quick! I just finished reediting my post to clean up all the errors, it should be good now.

I'm going to bump this once, hopefully someone can give me some feedback on it. If further clarification on the problem is required, don't hesitate to ask, thanks.

Hey,
I hope this isn't too late for your (my?) assignment. First of all, I would suggest using the form (a+xn)/(xn+1); in my opinion it's easier to work through and see the form. Second, the easiest way I know of to solve this question requires a little lemma. Write Xn+1 in terms of Xn, and if you begin with x1 you know the relationship between X1 and a so you can find the relationship between x2 and sqrt(a). [ie (a+x1)/(1+x1)=x2. What does x1= from this expression? you know when you get it to x1 you will have an expression >sqrt(a), and then solve again for x2.)

If you can see the relationship between x2 and sqrt(a), x3 and sqrt(a), x4 and sqrt(a)... and so on, and prove the general case (for even and for odd) then you will be able to approximate the form of xn+1 (if you write it in terms of Xn-1).

Hope that helps!

## What is the Sequence Problem in Rudin's PMA Ch.3 #17?

The Sequence Problem in Rudin's PMA Ch.3 #17 refers to a mathematical question that asks whether a given sequence of real numbers converges to a limit, and if so, what the limit is.

## What is the main goal of Rudin's PMA Ch.3 #17?

The main goal of Rudin's PMA Ch.3 #17 is to provide a rigorous and formal introduction to the concept of sequences and their convergence. This is an important foundation for many areas of mathematics, such as analysis and calculus.

## What is the solution to Rudin's PMA Ch.3 #17?

The solution to Rudin's PMA Ch.3 #17 involves applying the definition of convergence of a sequence, as well as various theorems and properties related to limits and sequences. It may also require using techniques such as the squeeze theorem or the Cauchy criterion.

## Can Rudin's PMA Ch.3 #17 be solved using other methods?

Yes, Rudin's PMA Ch.3 #17 can potentially be solved using other methods, such as the monotone convergence theorem or the Bolzano-Weierstrass theorem. However, the most direct and rigorous approach is usually through the use of the definition and related theorems.

## Why is understanding the Sequence Problem important in mathematics?

The Sequence Problem is important in mathematics because it is a fundamental concept that is used in many areas of the subject, including calculus, analysis, and algebra. It also helps to develop critical thinking and problem-solving skills, which are essential in all areas of mathematics and other fields as well.

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