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What is the sum of all positive integers divided by the square of the amount of positive integers?

$$\frac{1+2+3+4+5+6+\cdots}{(1+1+1+1+1+1+\cdots)^2}$$

Then the limit is \begin{align*} \lim_{n\to \infty}{ \frac{(1+2+3+\cdots+n)}{n^2}} & = \lim_{n\to\infty}{\frac{n(n+1)/2}{n^2}} \\ & = \lim_{n\to\infty}{\frac{n^2+n}{2n^2}} \\ & =\lim_{n\to\infty} \left( \frac{1}{2}+\frac{1}{2n} \right) \\ & = \frac{1}{2}. \end{align*} How would you do, for example, the sum of all real numbers between $0$ and $1$ inclusive, divided by the square of the amount of such numbers?

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3  
You wouldn't. These sums are not well-defined. See math.stackexchange.com/questions/70194/… . –  Qiaochu Yuan Jul 22 '12 at 2:08
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There is no such thing as an "infinite sum"; we can only do finite sums. What we do is take the finite partial sums, and then call the limit of this "the infinite sum", though we never really add up infinitely many things. You can't really do that with uncountably many terms; the closest analog would be an integral, perhaps $$\int_{0}^{1}\frac{x}{\int_0^x y^2\,dy}\,dx$$ –  Arturo Magidin Jul 22 '12 at 2:08
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@Michael: you're misreading the question. Read the last sentence. –  Qiaochu Yuan Jul 22 '12 at 2:23

4 Answers 4

up vote 4 down vote accepted

As Arturo Magidin notes in the comment to your question, the closest analog (that gives non-trivial answers) to such sums are usually integrals.

This is because an integral, intuitively, measures how much mass certain subsets of the real numbers have. In the text that follows I shall examine one such interpretation that comes to mind and try to motivate is as clearly as I can.

So, we interpret "sum of all numbers on $[0,1]$" to instead mean something like the mass of all points in $[0,1]$, assuming "each point has a mass proportionate to its value". Actually the masses of the points are, in a non-rigorous sense, infinitesimal, so we instead ask that this holds for the density, i.e. we assume that the density at the point $x$ is equal to $x$. This gives us the integral $\int_0^1x\,\mathrm{d}x$.

To calculate the "amount of" points in $[0,1]$, we repeat basically the same thought process. But as above, we were interested in the "density of value" which varied as the value varied, in this case we are interested in the "density of amount". So in the case above, the "mass" we were interested in was the value. In this case the "mass" we are interested is amount. Now in contrast with value, amount is spread evenly along the interval $[0,1]$. So the density function is constant in this case. We take it to be equal to $1$ everywhere, to assign the interval $[0,1]$ a mass of $1$, which seems to make sense. This gives us the integral $\int_0^1\,\mathrm{d}x$.

So, with this interpretation, the answer to your question would be $$\frac{\int_0^1x\,\mathrm{d}x}{(\int_0^1\mathrm{d}x)^2}=\frac{\frac12}{1^2}=\frac12,$$ which, interestingly, gives the same answer you obtained for your first question.

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Nice answer. The OP is clearly looking for a way to formalize an intuitive notion, so answers saying that his/her question is ill defined are not helpful. The OP's distinction between countable and uncountable should actually be posed as a distinction between discrete and continuous. Although we think of an integral as being over some uncountable subset of the real line, the Riemann sum is defined purely in terms of countable sets. The notion of countable versus uncountable was not even invented until hundreds of years after integration was invented. –  Ben Crowell Jul 23 '12 at 0:00
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@Ben: To be fair Newton and Leibniz were working in non-standard analysis, so they had to know some model theory and therefore enough set theory to discern countable from uncountable... :-) –  Asaf Karagila Jul 23 '12 at 0:02

Your question as posed doesn't make too much sense. But if you tighten it up and try to ask it in a more rigorous way, the answer's going to be zero.

The reason that the answer will come out zero is that if you just choose $n$ of the real numbers between 0 and 1, and try to evaluate the fraction, you'll have something like $\frac{x_1+x_2+\cdots+x_n}{n^2}$; but since each of the $x_i$ is less than 1, this fraction is less than $1/n$. As you put more and more numbers in, $n$ increases; so this fraction can be forced to be less than any given $\epsilon > 0$, and since it has to be positive, this proves that the answer is zero.

However, to speak of "the sum of an infinite number of real numbers, divided by some other infinite quantity" really doesn't make any sense at all. To try and show you why, consider your original question, involving the positive integers. You could equally well have written the sum as $\frac{1+2+3+4+\cdots}{((1+1)+(1+1)+(1+1)+(1+1)+\cdots)^2}$, and tried to work out this limit instead. It's the same sum, but the limit would come to 1/8 instead of 1/2.

When asking any questions about infinite quantities, you should check very carefully whether the thing you're asking for is really defined. If you're discussing dividing one infinite quantity by another, it's probably not. Sure, you can redefine things in terms of limits, but it's not clear how to do this when you start counting quantities that aren't actually countable.

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Your first expression is the same as $$ \frac{\lim\limits_{n\to\infty} 1+2+3+\cdots+n}{\lim\limits_{n\to\infty}(1+\cdots+n)^2}.\tag{1} $$ Your second is $$ \lim_{n\to\infty}\frac{1+2+3+\cdots+n}{(1+\cdots+1)^2}.\tag{2} $$

The second limit exists and is equal to $1/2$. The limits in the first expression do not exist, unless one takes them each to be $\infty$, in which case one cannot do arithmetic.

One sees things like $$ \frac{\lim f}{\lim g} = \lim\frac f g,\tag{3} $$ provided the two limits on the left side exist. Notice the word "provided". If they don't exist, then equalities like that in $(3)$ are not true. And "exist", in this context, would mean they are finite real numbers (and in the case of quotients, that the limit in the denominator is not $0$).

However, there are things like nonstandard integers in nonstandart analysis, and if one took $n$ to be one of those, then your first fraction would not be exactly $1/2$, but would differ from $1/2$ by an infinitely small nonstandard real number.

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Be careful with infinite limits of partial sums. What if I listed the naturals in the numerator like this?

$1,2,4,3,8,16,5,32,6,64,7,128,9...$

listing $n$, then $2^n$, omitting numbers if I've previously written them. It's quite clear that I'll hit every number this way, but the limit of sums is obviously not going to be $1/2$, or converge at all, since the top sum rises exponentially.

For infinite sums you can get away with a lot, since you can rearrange infinitely many numbers, pushing some numbers arbitrarily far back and changing your limit of partial sums. The key point is this: infinite limits of partial sums can depend on the order of terms. By the definition of uncountable, you can't list the terms, in any kind of order. So such a limit of partial sums isn't even defined. If you can't pick an order of terms, you haven't even picked a problem at all.

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