A series proof problem:

[cosine]

If you say "add all the real numbers from a to b," you would get an infinite answer, but if you say "add all the intergers from a to b" you get a finite answer. The problem I pose to you SFNers is:

Prove that for any a and b that are not the same, prove that the infinite series summing all the rational numbers between them diverges.

(Hint: You don't need to sum all of the rational numbers between a and b if you can show a series with only some of terms diverges.)

 

Edit: BTW, just to let you know, this is a little challenge problem I'm giving to the community that arised from a conversation between ecoli and me. Just so you don't think its hw or anything.

[ecoli]

that was quick!

[cosine]

that was quick!

I copied and pasted it from our conversation.

[matt grime]

Unless you specify some total ordering of the numbers there is no canonical way to add up the infinite set. Addition doesn't work like that.

 

You even rely on picking a series that diverges which is false on two counts: firstly there is no ordering so there is no such thing as a series, and secondly

 

1-1/2+1/3-1/4+1/5-...

 

converges to log 2, yet it has divergent subseries, moreover rearranging terms allows me to sum that puppy to log3/5 or something.

 

 

 

And the situation of adding "from" -a to a demonstrates that I can pick a total ordering which has a vaguely meaningful transfinite sum, quite possibly to almost any limit I wanted.

 

Presumably you were thinking of strictly positive numbers, and adding an infnite number of them which do not converge to zero.

[Lyssia]

Unless you specify some total ordering of the numbers there is no canonical way to add up the infinite set.

 

But the rationals are countable, right? A countable set isn't necessarily well-ordered?

 

Hm, but the rationals are dense.

 

Don't mind me, I forgot all my maths soon after I graduated.

[matt grime]

Any set is well ordered (or can be; it is not required to agree with the ordering of < that is given on the reals) if we assume the axiom of choice. How do you add the series x_n=1/n^2 and get pi^2/6? By adding the first term and the second and so on, and I'm certainly adding a countably infinite set....

 

the point is if you say to me "add up these numbers" i am entitled, nay necessitated, to say "what is the first, second, third term" etc, ie ask the order of summing as my example shows. it makes no sense otherwise. or to put it another way, no matter what the two people discussing this first thought, the question as asked is completely meaningless. I am asking them to demonstrate to me that they know how to sum something that is not either a finite sum or a series, because it is not an obvious thing to be able to do.

[cosine]

Any set is well ordered (or can be; it is not required to agree with the ordering of < that is given on the reals) if we assume the axiom of choice. How do you add the series x_n=1/n^2 and get pi^2/6? By adding the first term and the second and so on' date=' and I'm certainly adding a countably infinite set....

 

the point is if you say to me "add up these numbers" i am entitled, nay necessitated, to say "what is the first, second, third term" etc, ie ask the order of summing as my example shows. it makes no sense otherwise. or to put it another way, no matter what the two people discussing this first thought, the question as asked is completely meaningless. I am asking them to demonstrate to me that they know how to sum something that is not either a finite sum or a series, because it is not an obvious thing to be able to do.[/quote']

 

The entire reason I thought the question was interesting in the first place is because it asked you to evaluate a sum with no evident starting point. I had some arguements, but I realize that parts of them are faulty:

If a and b are either both positive or both negative, the problem reduces to first showing that between any two distinct reals there are two distinct rationals. And then showing between any two distinct rationals there are an infinite amount of rationals between them. Thus you sum a series whose terms do not approach zero.

If a and b are of opposite signs, then the fault in the question is that the result depends how you sum the series, since it is not absolutely convergent (based on the case above). So the question in this case is meaningless without a prescribed order. Sorry world.