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## Main Question or Discussion Point

If I added all the numbers between 0 and 1, what do I get?

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If I added all the numbers between 0 and 1, what do I get?

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LeonhardEuler

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- #3

TD

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If you're talking about the reals, the answer is indeed [itex]\infty[/itex]

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[tex] \sum^{\infty}_{i =1} \frac{1}{i} [/tex] is a divergent series, so it has to be true. Seems that it is not something easy to think about intuitively

- #5

Icebreaker

Either adding up "all" the reals or rationals between 0 and 1 will produce infinity

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It seems strange that for p=1, the p-series is divergent. Why is this so? I mean, it seems like each term will be smaller than the one before it, and [itex]\lim_{n\to\infty}\frac{1}{n}=0[/itex] so it (at least intuitively) should be convergent.TOKAMAK said:

[tex] \sum^{\infty}_{i =1} \frac{1}{i} [/tex] is a divergent series, so it has to be true. Seems that it is not something easy to think about intuitively

Any thoughts on this?

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[tex] \int^{\infty}_{a} dx \frac{1}{x} [/tex]

when integrated gives

[tex] \ln{x}|^{\infty}_{a} = \ln{\infty} - \ln{a} = \ln{\frac{\infty}{a}} [/tex]

But infinity divided by a constant is just infinity, and the natural log of infinity isn't bounded either. So, I guess through the use of the integral test we can indeed establish that the series diverges. To be honest, I don't remember if there was in fact a reason given in my calc class as to why the integral test works (it's been some time), so perhaps someone might have a better, or more in depth answer as to why the harmonic series diverges, and therefore, adding all the real numbers between, say, 0 and 1 is infinite.

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shmoe

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One way to see the divergence is to group the terms like so (this is a little sloppy how I'm dealing with the limit, but you can make everything precise):apmcavoy said:It seems strange that for p=1, the p-series is divergent. Why is this so? I mean, it seems like each term will be smaller than the one before it, and [itex]\lim_{n\to\infty}\frac{1}{n}=0[/itex] so it (at least intuitively) should be convergent.

Any thoughts on this?

1+1/2+1/3+1/4+...

=1+1/2+(1/3+1/4)+(1/5+1/6+1/7+1/8)+(1/9+...1/16)+...

>=1+1/2+(1/4+1/4)+(1/8+1/8+1/8+1/8)+(1/16+...+1/16)+...

=1+1/2+2*(1/4)+4*(1/8)+8*(1/16)+...

=1+1/2+1/2+1/2+...

so by adding enoiugh terms you can make the original series as large as we like.

The integral comparison test works fine as well and shows that the divergence is logarithmic (though you might have guessed that from the above). The integral test can be loosely justified by drawing the graph of 1/x and drawing a rectangle of hieght 1 from x=1 to 2, a rectangle of height 1/2 from 2 to 3, a rectangle of height 1/3 from 3 to 4, etc. The area of the first n rectangles is the sum of the first n terms of the harmonic series, and contains the graph of 1/x and hence the harmonic series diverges at least as fast as log. By shifting the rectangles one unit left you get the rectangles below the 1/x graph, so they will both have the same order of growth.

Incidently, if you have any uncountable set of positive real numbers, then you can find a sequence of (distinct) numbers in this set whose series diverges.

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arildno

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Zurtex

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I'm not quite sure that would make too much sense given the way the poster phrased the question.arildno said:

But yes, it is quite important to define what type of numbers you are talking about before you can prove anything.

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arildno

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I don't think this makes a lot sense, either.whozum said:If I added all the numbers between 0 and 1, what do I get?

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matt grime

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[1/n,1/(n-1))

each if the sum were in any sense to be finite then there can only be a finite number in each class, but then the set is a countable union of finite sets, hence countable, contradicting the assumption we were summing an uncountable set of numbers.

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That makes sense. It just didn't seem divergent intuitively.

Thanks.

Thanks.

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matt grime

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HallsofIvy

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The requirement that sequence a_{n} go to 0 in order that the sum

[tex]\Sigma_{n=0}^{\infty} a_n[/tex] converge is a**necessary** condition, not a **sufficient** condition.

[tex]\Sigma_{n=0}^{\infty} a_n[/tex] converge is a

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I think thats a weak assumption. On the real scale and numbers between 0 and 1 clearly means any real number between the two if you ask me.arildno said:

Thanks for everyones input.

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matt grime

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1. what order are you summing them in and

2. obviously it diverges and if you'd bothered to check if this question had been asked before in this forum (which it has, in some sense; there was a thread about uncountable indices and summation, if not two)

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Sorry I dont have a PhD in math, i dont know what an uncountable index is. I don't see why it wouldnt make sense, and I dont see why the order would matter.matt grime said:

1. what order are you summing them in and

2. obviously it diverges and if you'd bothered to check if this question had been asked before in this forum (which it has, in some sense; there was a thread about uncountable indices and summation, if not two)

My question was answered, I said thanks, what more do you want?

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matt grime

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Irrespective of the order issue, the set you want to sum over contains only positive terms, and one ordering of them would be to sum 1/2, 1/3, 1/4 etc which is known to diverge.

We know how to add up a finite number of terms:

x+y+z,

we know how to take the limit to a set indexed by the natural numbers

x_1+x_2+x_3+.....

but how were you going to add up all the numbers in [0,1]? What limit were you going to take? This is why, to me, it "clearly doesn't make sense" since there is no indication given of how to actually start summing this series. (the infinite set of numbers between 0 and 1 cannot be indexed by the natural numbers and is thus uncountable, that is the definition of uncountable).

You said, rudely in my opinion, that arildno made a weak assumption, hence my (rude) reply; I'd say he made the most sensible assumption of them all.

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arildno

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I assumed that you asked a sensible question. If that's a weak assumption, so be it.whozum said:I think thats a weak assumption. On the real scale and numbers between 0 and 1 clearly means any real number between the two if you ask me.

Thanks for everyones input.

- #22

jcsd

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Any sequence whose set of terms are anywhere dense will diverge.

I don't know how widely known these two staments are as I thoguht of them and proved themself, but as they only require a basic amount of analysis to prove doubt I am the first person to notice this general property of these classes of sequences and sums.

- #23

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Just take a whole bunch of high numbers and start adding them:

.9 + .99 + .999 + .9999 + .99999 + ...

You're nearly adding one every time. When do you want to stop? And you still have .8, .98, .998, etc. left to add to that :P

...

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jcsd

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As has alreday been pointed out the infinite sums are only defiend on sequences whoses set fo terms is countable, but we could for example, sum all the rationals in the interals (0,1). Howvere we arranged ththsi set into a sequence of terms the sequnece of partial sums would always diverge.Moo Of Doom said:

Just take a whole bunch of high numbers and start adding them:

.9 + .99 + .999 + .9999 + .99999 + ...

You're nearly adding one every time. When do you want to stop? And you still have .8, .98, .998, etc. left to add to that :P

...

- #25

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You have thejcsd said:.. .. for example, sum all the rationals in the interals (0,1). Howvere we arranged ththsi set into a sequence of terms the sequnece of partial sums would always diverge.

I'd hate to beat up a dead horse, but can't the sum of all rationals in [tex] [0,1] [/tex] just be represented as

[tex]\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 0}^n k \, , \; \text{where} \; n \in \mathbb{N} [/tex]

--------------------------------------------------------------------------

And so it thus diverges :

[tex]\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 0}^n k = \mathop {\lim }\limits_{n \to \infty } \frac{{n + 1}}{2} = \infty [/tex]

Even though you have the

subtracting 0 or 1 from my series changes nothing; it still diverges.

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