This is the Grandi's series: $1-1+1-1+1-1+\dots$

The series can be equal to $0$


or to $1$


or to $1/2$

$$S=1-1+1-1+1-\dots,\quad\quad S=1-(1-1+1-1+1-1+1-...)$$ $$\Rightarrow S=1-S\Rightarrow 2S=1\Rightarrow S=1/2$$

Isn't this a contradiction? The integers are closed under addition and subtraction, but we get a fraction. Why?

  • 3
    $\begingroup$ Related : math.stackexchange.com/questions/425226/… $\endgroup$ Jun 23, 2013 at 16:04
  • $\begingroup$ no, it is not a contradicition. closed under addition means that any finite sum of integer is an integer. Are you talking about finite sums of integers? What are you talking about? It is a contradiction of you say $S=0$ and $S=1$ because the numbers $0$ and $1$ are different in $\mathbb{Z}$ $\endgroup$
    – miracle173
    Feb 17, 2014 at 10:12
  • $\begingroup$ The problem is that in the proof of $S=0$ you assume that the number of the terms in the sum is even, whereas in $S=1$ you assume that it's odd. But infinity is not an even or an odd number (it's not an integer number at all). Also, in the proof that $S=1/2$ the mistake is that you assume that $S−1=S$ which is not true either. $\endgroup$
    – Andreas K.
    Jun 30, 2016 at 19:56

4 Answers 4


The issue here is that addition of a finite number of terms is a nice, well-defined operation; addition of an infinite number of terms need not be.

In particular, rearranging terms in an infinite sum can actually change the sum. The series you have presented does not converge - and so discussing it as having a value is incorrect to begin with. However, there are series which converge, and which STILL cannot be rearranged without changing the value: this is true for series that are called conditionally convergent - that is, the series converges, but the sum of absolute values of the terms does not.

Interestingly, for conditionally convergent series, it turns out that you can rearrange the terms to give any limit that you like.

At the end of the day, the big thing to remember is this: addition of an infinite number of terms cannot be assumed to be associative or commutative, unless you know that the sum of the absolute values of the terms converges - a property known as absolute convergence.


Just because you can write a string of symbols that appear to respect normal mathematical syntax, such as $1-1+1-1+1-1+\cdots,$ it does not follow that the string is meaningful. Many mathematicians would indeed regard a string like this as meaningless, and not bother with it. Some, though, have played formal games with such strings to prove them "equal" to any number.


$$S(5) = 1-1+1-1+1 = 1$$ then $1-S(5) = 1-1+1-1 = S(4)$. So $1-S(n) = S(n-1)$. If $S(5)=S(4)$ (which is not), then $S(5)=1/2$, wrong, and similar to $S=1/2$. Grandi assumes $S(\infty)=S(\infty-1)=S$, so that $S=1/2$.

The question is: Is it really $S(\infty)=S(\infty-1)$?. Of course $\infty-1= \infty$, but this is not the case, now $\infty$ is not a value, is the order of the variable which still will take values of $1$ or $-1$, even their order is $\infty$.

So $S(\infty)$ does not equal $S(\infty-1)$. That's what I believe.


Some old fashioned consistency should dispel La Grandi Llusion and reveal that no contradictory sums are obtainable. We start with a sequence 1,-1,1,-1,1,-1 and construct from it a series 1+-1+1+-1+1+-1..., and bracket it thus: (1+-1)+(1+-1)+(1+-1).... So far so good, the sum is 0. Now often we get people re-bracketing thus: 1+(-1+1)+(-1+1)+(-1+1)... But look, and see that an extra 1 has also been sneaked in, so no wonder the sum has changed to 1. The consistent and correct way to re-bracket is: 1+(-1+1)+(-1+1)+(-1... which gives 0 as before. To be more thoroughly consistent the hanging single bracket on the right: "(" should be balanced by its counterpart: ")" on the left after that first 1. We now have +1)+(-1+1)+(-1+1)+(-1 ... Those dots at the end now simply mean we can join it up with the beginning in a cyclic loop. You don't want a loop? OK, another re-bracketing of improved consistency goes: 1+(-1+1)+-1+(1+-1)..., again summing to 0. That infinite series business is a red herring which has hitherto been used as a cover for logical legerdemain.

  • 1
    $\begingroup$ This is a good example of illogical legerdemain. $\endgroup$ Jun 2, 2021 at 15:39

Not the answer you're looking for? Browse other questions tagged .