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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    **Every** real number is the sum of countably many rational numbers. – Did Mar 04 '12 at 09:47
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    Properties that are preserved under a finite number of operations are not necessarily preserved under an infinite number of them. –  Mar 04 '12 at 09:57
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    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. – Henry Mar 04 '12 at 10:15
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    because real number are closure of rational numbers. – quartz Mar 04 '12 at 12:12
  • @RahulNarain A nice example is how [we can infinitely divide by two to divide by three](http://math.stackexchange.com/questions/116987/is-it-possible-to-get-1-3-without-dividing-by-3/116991#116991). – Pedro Mar 07 '12 at 02:32
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    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 $\prod_{n=1}^\infty 0.5 = 0$ – leonbloy Apr 30 '12 at 18:36

7 Answers7


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...

Mariano Suárez-Álvarez
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    Interesting... as a side note, you can determine the size of an infinite set by seeing if it can fit inside another infinite set. This isn't directly relevant here, but still is something important to understand. http://en.wikipedia.org/wiki/Aleph_number – JSWork Mar 05 '12 at 15:06
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    @JSWork, not only would I say that that isn't directly relevant here: it is quite irrelevant, in fact! :D – Mariano Suárez-Álvarez Mar 07 '12 at 04:11
  • @MarianoSuárez-Álvarez Is infinite sum of rationals always an irrational number ? Because recently I saw an example that the infinite sum of 1/2n for n from 1 to infinity equals 1, and 1 is rational. – Maths Survivor Dec 20 '17 at 23:14
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    @MathsSurvivor No. Your own comment has the proof. – user3658307 May 11 '18 at 18:37

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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    Good point to observe! –  Mar 04 '12 at 10:25
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    You *can* add an infinite sequence of rational numbers, under certain conditions. This is a cornerstone of analysis. – TonyK Mar 05 '12 at 00:30
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    @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. – Mariano Suárez-Álvarez Mar 05 '12 at 02:06
  • @Mariano, yes, you really can. Real analysis gives a meaning to the infinite sum. – TonyK Mar 05 '12 at 18:31
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    @TonyK, can you point to any reference where an 'infinite sum' is considered? – Mariano Suárez-Álvarez Mar 05 '12 at 18:41
  • @Mariano: Rudin's Principles of Mathematical Analysis (for instance) explains how to give meaning to certain infinite sums of rationals. – TonyK Mar 05 '12 at 19:56
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    @TonyK, Dear TonyK: if you read Rudin with care you will see that it nowhere does that. – Mariano Suárez-Álvarez Mar 05 '12 at 20:05
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    @Mariano: One of the aims of real analysis is to give a meaning to infinite sums. I'm surprised you don't know this. And yes, I read Rudin with great care in 1977, so you don't have to patronise me, dear Mariano. – TonyK Mar 05 '12 at 20:37
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  • Infinite sums have meaning, sure. But I have a finite lifetime, and each addition step takes a finite time -- therefore I cannot add infinitely many numbers. Neither can a computer, by the same reasoning. However I *can* use reasoning in some cases to compute the limit of an infinite sum (as could a computer in some other cases, with sufficient programming). The difference may be a bit pedantic, but this is mathematics, after all. – Jason S Mar 07 '12 at 15:52
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    @TonyK $$\sum_{n=1}^\infty a_n \stackrel{\text{def}}{\equiv}\lim_{k\rightarrow \infty} \sum_{n=1}^k a_n$$ –  Sep 06 '12 at 07:43
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    @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." – Pedro Apr 15 '13 at 19:31
  • This talk of what you can and can't do seems pretty meaningless. Yes, infinite sum is defined as a limit of finite sums, when such a limit exists. That's *the definition* of infinite sum, and yes, it's shorthand. If you're trying to point out something else (maybe that there are perils in assuming the limit always exists?), than maybe say that, instead of asserting we "can't do" something which we certainly can do, which we do all the time, and which is useful. – Don Hatch Jul 30 '18 at 08:05

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 $\mathbb Q$.

Neil Peterman
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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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    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. – Rafid Mar 04 '12 at 10:41
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    @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. – Paul Mar 04 '12 at 15:00
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    @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. –  Mar 04 '12 at 22:31

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 $$e = \sum_{k=0}^{\infty}{\frac{1}{k!}} $$


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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