For example how come $\zeta(2)=\sum_{n=1}^{\infty}n^{-2}=\frac{\pi^2}{6}$. It seems counter intuitive that you can add numbers in $\mathbb{Q}$ and get an irrational number.
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101$\begingroup$ Every real number is the sum of countably many rational numbers. $\endgroup$– DidCommented Mar 4, 2012 at 9:47
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56$\begingroup$ Properties that are preserved under a finite number of operations are not necessarily preserved under an infinite number of them. $\endgroup$– user856Commented Mar 4, 2012 at 9:57
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7$\begingroup$ Real numbers have the property that the bounded infinite sum of positive real numbers is a real number because of completeness. This is not a property of rational numbers, so I would intuitively expect some bounded infinite sums of positive rational numbers not to be a rational number. $\endgroup$– HenryCommented Mar 4, 2012 at 10:15
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4$\begingroup$ because real number are closure of rational numbers. $\endgroup$– quartzCommented Mar 4, 2012 at 12:12
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3$\begingroup$ Another example to illustrate Rahul's point: the product of infinite positive reals can yield zero: $\prod_{n=1}^\infty 1/n = 0$ or more simply $\prod_{n=1}^\infty 0.5 = 0$ $\endgroup$– leonbloyCommented Apr 30, 2012 at 18:36
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7 Answers
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But for example $$\pi=3+0.1+0.04+0.001+0.0005+0.00009+0.000002+\cdots$$ and that surely does not seem strange to you...
answered Mar 4, 2012 at 9:45
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You can't add an infinite number of rational numbers. What you can do, though, is find a limit of a sequence of partial sums. So, $\pi^2/6$ is the limit to infinity of the sequence $1, 1 + 1/4, 1 + 1/4 + 1/9, 1 + 1/4 + 1/9 + 1/16, \ldots $. Writing it so that it looks like a sum is really just a shorthand.
In other words, $\sum^\infty_{i=1} \cdots$ is actually kind of an abbreviation for $\lim_{n\to\infty} \sum^n_{i=1} \cdots$.
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24$\begingroup$ @TonyK, nop, you really cannot. What you can do is to take the limit of sequences of partial sums: there is a difference between that and «summing an infinite sequence of numbers». In particular, the latter simply does not make sense. $\endgroup$ Commented Mar 5, 2012 at 2:06
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9$\begingroup$ @TonyK, Dear TonyK: if you read Rudin with care you will see that it nowhere does that. $\endgroup$ Commented Mar 5, 2012 at 20:05
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4$\begingroup$ @TonyK $$\sum_{n=1}^\infty a_n \stackrel{\text{def}}{\equiv}\lim_{k\rightarrow \infty} \sum_{n=1}^k a_n$$ $\endgroup$– user38268Commented Sep 6, 2012 at 7:43
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31$\begingroup$ @TonyK Quoting from Rudin: "The number $s$ is called the sum of the series, but it should be clearly understood that $s$ is the limit of a sequence of sums and is not obtained simply by addition." $\endgroup$– Pedro ♦Commented Apr 15, 2013 at 19:31
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Others have demonstrated some examples that make clear why this can happen, but I wanted to point out the key mathematical concept here is "Completeness" of the metric space. A metric space is any set with "distance" defined between any two elements (in the case of $\mathbb{Q}$, we would say $d(x,y) = |x-y|$). A sequence $x_i$ is "Cauchy" if late elements stop moving around very much, a necessary condition for a sequence to have a finite limit. To put it formally, ${x_i}$ is cauchy for $\epsilon>0$, there is a sufficiently large $N$ so that for every $m,n>N$ we have $d(x_n,x_m)<\epsilon$. A metric space is complete if all Cauchy sequences have a limit in the space. The canonical complete metric space is $\mathbb R$, which is in fact the completion of $\mathbb{Q}$, or the smallest complete set containing $\mathbb Q$.
We think of an infinite sum as the limit of a sequence of partial sums: $$\sum_{n=1}^\infty x_n = \lim_{N\to\infty}\left( \sum_{i=1}^Nx_n \right)$$ As others have pointed out with a number of good counter-examples (my favorite of which is the decimal representation of an irrational number), $\mathbb{Q}$ is not complete, therefore an infinite sum of elements of $\mathbb Q$, for which partial sums are necessarily elements of $\mathbb Q$, can converge to a value not in $\mathbb Q$.
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It is counter-intuitive only if you are adding a "finite" number of rational numbers. Otherwise, as @Mariano implied, any irrational number consists of an infinite number of digits, and thus can be represented as a sum of rational numbers.
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2$\begingroup$ This is my point. There is nothing magic with irrational numbers and they just use the same digits that rational numbers, so there is no surprise or counter-intuition in having irrational numbers representable as an infinite sum of rational numbers. $\endgroup$– RafidCommented Mar 4, 2012 at 10:41
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2$\begingroup$ @RahulNarain: In the decimal expressions of rational numbers, there is always either a terminating or an infinitely repeating string of digits. Irrational numbers have neither of these properties, but can still be expressed as an infinitely long non-repeating sequence digits. Though there may not be an explicit formula for this sequence, it can still be thought of as an infinite sum of rational numbers, as Rafid indicated. $\endgroup$– PaulCommented Mar 4, 2012 at 15:00
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1$\begingroup$ @Paul, I was making a joke on the phrase "must have digits from 0 to 9" that appeared in the original version of the answer. $\endgroup$– user856Commented Mar 4, 2012 at 22:31
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Besides that. Given a sequence of positive rational numbers such that their sum converges and such that $a_n > \sum_{k = n + 1}^\infty a_k$. Then choosing a $\pm$ sign for each term of the sequence gives a new convergent series, each to a different number. By a countable-uncountable argument you get nonnumerable examples of that kind of series :).
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For example look at $$e = \sum_{k=0}^{\infty}{\frac{1}{k!}} $$
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This has to see with the rate at which the sum/series converges to its limit and the Roth-Thue-Siegel theorem which allows you to use the rate of convergence to decide if the limit is rational or not.
Maybe Emile (the OP) meant to ask something of this sort (please let me know, Emile): why do some (convergent, of course) infinite sums of rationals are rational and others are irrational?
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$\begingroup$ How so? The conditions/rate of convergence affect whether the limit is rational or not. Don't they? $\endgroup$– AQPCommented Mar 6, 2012 at 16:54