Numbering the Rationals for Convergence for Infinite Sum Calculation

[lukepeterpaul]

hi, i would be very grateful if someone could start me off me in the right direction in numbering the rationals such that the infinite sum of (x_n - x_n+1)^2 (where x_n are rationals) converges.
thanks!

 

[Yayness]

Could you be more specific? This alone is not an infinite sum:
$$(x_n-x_{n+1})^2$$

Is it one of the terms in the sum? What exactly does the sum look like? Is there a relation between x_n and x_n-1 and x_n-2 etc?
Is x₁ (Or eventually x₀) described in any way?

Could you describe the problem more detailed?

 

[lukepeterpaul]

$$\sum(x_n - x_n+1)^2$$ over n, where n ranges from 1 to infinity
this is all the info I've got...

 

[lugita15]

$$\sum(x_n - x_n+1)^2$$ over n, where n ranges from 1 to infinity
this is all the info I've got...

I think you mean $$\sum (x_n-x_{n+1})^2$$.

 

[bpet]

If repeats are allowed, try breaking up the sequence into a number of blocks where for the kth block the difference between successive terms is small, say 1/k^4. Say block 1 goes from 0 to 1, block 2 goes from 1 to -2, block 3 goes from -2 to 3, etc (so the sequence is 0,1,1-1/16,1-2/16,...,-2,-2+1/81,...; the contribution of the kth block to the sum of squares is (2k-1)*k^4/k^8 so the overall sum converges (and every rational is visited infinitely often).

 

[disregardthat]

hi, i would be very grateful if someone could start me off me in the right direction in numbering the rationals such that the infinite sum of (x_n - x_n+1)^2 (where x_n are rationals) converges.
thanks!

Ok, here is my idea: Take any ordering of the rationals y_1,y_2,... You can recursively weed out duplicates, so we assume this.

If you just want a tip: Try to cut up each interval $$(y_n,y_{n+1})$$ with increasingly fine intervals in an appropriate way, and extend the ordering with the cut points between the y_n's, and find an upper bound of the sum. And don't read further.

If you are stuck:
For each n: If $$y_n<y_{n+1}$$, define $$z_{kn}=y_n+2^{-n}\sum^k_{i=1}\frac{1}{i}$$, for k ranging from 1 to the smallest K_n such that $$y_{n+1}-z_{K_nn} \leq 2^{-n}$$, K_n exists since $$\sum^{N}_{i=1}\frac{1}{i}$$ grows without bound, and $$z_{k+1n}-z_{kn} \leq 2^{-n}$$. If $$y_n>y_{n+1}$$, then do the same by switching the plus-sign to a minus-sign in from of the $$2^{-n}$$*etc when defining $$z_{kn}$$. What we are doing is chopping up the interval $$(y_n,y_{n+1})$$ into small parts by the $$z_{kn}$$'s.

Now define $$x_1 = y_1, x_2 = z_{11},x_2=z_{21}, ..., x_{K_n+1}=z_{K_n1},x_{K_n+2}=y_2, x_{K_n+3}=z_{12}$$, and so on. We are enumerating the rationals by picking each $$y_n$$ and then each consecutive $$z_{kn}$$ until we reach the last, then continuing similarly with $$y_{n+1}$$.

Now, find (an upper bound of) the sum!

 

[disregardthat]

Some interesting follow-up questions:

Can the sum exist if the x_n's are all distinct?
Can the sum over |x_n-x_{n+1}| exist (distinct or not)?

 

[dimitri151]

The sum can not converge as I've understood the problem. If you include the integers then there will be an infinite number of terms which are greater than or equal to one and the series can't converge. Do you mean the rationals in (0,1) or something like that?

 

[disregardthat]

The sum can not converge as I've understood the problem. If you include the integers then there will be an infinite number of terms which are greater than or equal to one and the series can't converge. Do you mean the rationals in (0,1) or something like that?

This is wrong, you could order the rational in such a way that the sum converges. Try to sum up the rationals ordered as I have defined.

 

[disregardthat]

I might as well post my saved calculations which I decided to hold. It seems OP is no longer here and it might be interesting for others.

Using the previous definition of x_n :
$$\sum_n(x_n-x_{n+1})^2 = \sum_n (y_n-z_{1n})^2+\left( \sum_{k=1}^{K_n-1}(z_{kn}-z_{k+1n})^2 \right) +(z_{K_nn}-y_{n+1})^2$$
$$\leq \sum_n (2^{-n})^2+\left( \sum_{k=1}^{K_n-1}(\frac{2^{-n}}{k+1})^2 \right) +(2^{-n})^2=\sum_n 2^{-2n+1}+2^{-2n}\sum_{k=1}^{K_n-1}\frac{1}{(k+1)^2}$$
$$\leq \sum_n 2^{-2n+1}+2^{-2n}\sum_{k=1}^{\infty}\frac{1}{(k+1)^2} = \sum_n 2^{-2n+1}+2^{-2n}(\frac{\pi^2}{6}-1)$$
$$=\sum_n 2^{-2n}(\frac{\pi^2}{6}+1) \\ =\frac{\frac{\pi^2}{6}+1}{1-2^{-2}}=\frac{2\frac{\pi^2}{3}+4}{5}$$

 

[lukepeterpaul]

jarle, thanks for your solution. however, would i be right in thinking that the proposed way of enumerating the rationals does not cover all the rationals? e.g. y1+1/1000, say, won't be included...? i apologize if I'm mistaken...

 

[disregardthat]

jarle, thanks for your solution. however, would i be right in thinking that the proposed way of enumerating the rationals does not cover all the rationals? e.g. y1+1/1000, say, won't be included...? i apologize if I'm mistaken...

By assumption y_1,y_2, ... is a numeration of the rationals, which can be defined in various ways. The most common one I believe is considering every integer lattice point in the plane. Since x_1,x_2,... contains y_1,y_2,... by construction, the sequence will contain y_1+1/1000 (there is an N such that y_N=y_1+1/1000), and any other rational number.

(I spot a slight typo while I was defining the sequence x_1,x_2,... Of course I mean K_1 where I wrote K_n)

 

[lukepeterpaul]

ah, so double counting is allowed
thanks again, that is very helpful
i've got another question: "given that sum x_n converges, where x_n are real, does sum (x_n)^3 necessarily converge?
my gut feeling is "no" when considering conditional convergent series. it may be that the cubing can increase the ratio of consecutive "groups of terms" (ie the terms in the series we consider as one during the ratio test) s.t. the series is no longer convergent.
m i right?
if so, how might i go about constructing an example?
thanks!