- #1
whozum
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If I added all the numbers between 0 and 1, what do I get?
TOKAMAK said:Yeah weird to think about I guess, but
[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
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?
I'm not quite sure that would make too much sense given the way the poster phrased the question.arildno said:If you include 0 and 1, then your sum would be 1. (I assume that with "numbers", you meant the integers).
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?
arildno said:If you include 0 and 1, then your sum would be 1. (I assume that with "numbers", you meant the integers).
matt grime said:*cough* and equally clearly asking what the sum of all numbers between 0 and 1 is makes so much sense..
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)
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.
Moo Of Doom said:Maybe a simple way to relate that the summation of all reals between 0 and 1 is divergent is this:
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
...
You have the open interval [tex] (0,1) [/tex].jcsd 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.
Indeed, now I find that to be true (after more careful reasoning), which is why, as I mentioned :shy:jcsd said:edited to add: I see you've edited your sum, the limit is now infty, but the ste of terms isn't the rationals in the inetrval [0,1] (as what you've got there isn't a sum to infinity).
By the "dead horse" :shy:, I'm referring to https://www.physicsforums.com/showthread.php?t=81526bomba923 said:I'd hate to beat up a dead horse
Tide said:I think whozum has an obligation to specify what number comes next after 0 if he/she expects any help! :)
matt grime said:Well, it isn't a clear cut question. The first thing to have asked yourself is: *how* do you try to sum all the numbers between 0 and 1. That is to say that there really was a more basic question going on. For instance it *is* possible to sum more than a simple series indexed by the natuarl nubmers.
eg, let w be a symbol and r run from 1 to infinity, then define x(w)_r to be a geometric series of positive terms that sums to 1/2, then x(2w)_r be one that sums to 1/4, then x(3w)_r be oen that sums to 1/8, and so on so that the sequence x(nw)_r sums to 1/2^n
then i claim that this is gives a series indexed by the ordinal w^2 (or something like that) whose sum which we do transfinitely is 1.
However [0,1] is uncountable. so we can't do this. So, it is perfectly reasonable to ask how you thought the sum was going to be taken.