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Grandi's series is defined as:

$$\sum_{n=0}^{\infty} (-1)^n = 1 - 1+1-1+\cdots$$

By plainly looking at this series it seems like the value of it is either $1$ or $0$ by doing the following groupings:

$$(1-1)+(1-1)+(1-1)+\cdots=0+0+0+\cdots=0$$

OR

$$1+(-1+1)+(-1+1)+\cdots=1+0+0+\cdots=1$$

However, if we say

$$A = 1 - 1+1-1+\cdots$$

Then

$$1-A = 1-(1 - 1+1-1+\cdots)=1 - 1+1-1+\cdots = A$$

and thus $A = 1/2$

We know that $\frac{1}{1+x} = 1 - x + x^2 - x^3 + \cdots$ which, when $x=1$, evaluates the sum as $1/2$ once again.

One final proof that the series converges to $1/2$ is as follows (similar to the above proof): $$1/2=\frac{1}{1+1} = 1-\frac{1}{1+1} = 1-\left(1-\frac{1}{1+1}\right)=1-\left(1-\left(1-\left(1-\left(\cdots\left(1-\frac{1}{1+1}\right)\right)\right)\right)\right) = 1-1+1-1+\cdots$$

This series apparently confused many of the great minds in math. Many, many more methods are known that also show that this sum is equal to $1/2$.

Why do many seemingly divergent sums such as this one seem to converge? Some other examples that can be evaluated include $1 − 2 + 3 − 4 + \cdots = -1/4$, $1 − 2 + 4 − 8+ \cdots = 1/3$, $1 + 1 + 1 + 1 +\cdots=-1/2$, $1 + 2 + 3 + 4 +\cdots=-1/12$, etc.

See http://en.wikipedia.org/wiki/Grandi's_series for more information about this series.

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    $\begingroup$ To someone who knows the definition of convergence of an infinite series, it doesn’t seem to converge. $\endgroup$
    – Brian M. Scott
    Commented Apr 9, 2012 at 1:40
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    $\begingroup$ $1-x+x^2-x^3+\dots$ converges iff $|x|<1$, in which case it converges to $\frac{1}{1+x}$. Plugging in $x=1$ to both sides doesn't make sense, because the equation no longer even makes sense (since the series side, as you've seen, is not a well-defined value). $\endgroup$
    – William
    Commented Apr 9, 2012 at 1:47
  • $\begingroup$ It would seem that way. However, the Wikipedia page suggests otherwise. Some more info on this series can be found here: en.wikipedia.org/wiki/History_of_Grandi%27s_series $\endgroup$
    – Argon
    Commented Apr 9, 2012 at 1:52
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    $\begingroup$ The answer to the question taken at face value is that the series do not converge under the standard definition of convergence of a series, but there are alternative ways to define the sum of a divergent series that do give a (possibly reasonable) finite answer. What I don't understand is your phrasing in the sentence "Why do many seemingly divergent sums such as this one seem to converge?" Which one is it, does it seem to diverge or does it seem to converge to you? $\endgroup$
    – user856
    Commented Apr 9, 2012 at 1:57
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    $\begingroup$ It diverges. There is no real number $x$ such that for each $\epsilon>0$ there is an $n_\epsilon\in\Bbb N$ such that $\left|x-\sum_{k=0}^n(-1)^k\right|<\epsilon$ for all $n\ge n_\epsilon$. $\endgroup$
    – Brian M. Scott
    Commented Apr 9, 2012 at 2:04

1 Answer 1

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For series with summands which do not have constant sign, one loses associativity of addition. Hence, parenthesizing before adding is not allowed. Furthermore, most of these "proofs" of what the value must be assume that the number exists and is real. If it does not exist, then all the algebra you do on the symbol is meaningless.

I think you might be interested in the Riemann Rearrangement Theorem.

It says that if you have a conditionally convergent series, that is $\sum a_i$ converges, but $\sum|a_i|$ diverges, then for every $\ell \in [-\infty,\infty]$ there is a reindexing $a_{i_k}$ of the $a_i$ such that $\sum a_{i_k} = \ell$. More simply, by adding the numbers in a different order, you can make the sum add up to anything you like (even $\pm \infty$).

This really drives home the point that when adding infinitely many terms, one must be much more careful about which properties of real numbers are preseved, and which are not.

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  • $\begingroup$ Sure thing! I thought of one other thing you might be interested in. You might ask the question "Why do we only consider adding countably many terms? Is it possible to add up uncountably many? Does $\sum_{\lambda \in \mathbb{R}} a_{\lambda}$ make sense?". It can be shown that for non-negative terms $a_{\lambda}$, $\lambda \in \Lambda \subseteq \mathbb{R}$, the sum $\sum_{\lambda \in \Lambda} a_{\lambda}$ can be finite only if there are only countably many nonzero terms, i.e. that it can be reduced to a series $\sum_{i=1}^\infty a_{\lambda_i}$. Just an interesting little factoid. $\endgroup$
    – nullUser
    Commented Apr 9, 2012 at 2:19
  • $\begingroup$ Once again, thank you! This seems to be the solution to what was stumping Grandi, Marchetti, Jacob and Daniel Bernoulli and Euler all those years. $\endgroup$
    – Argon
    Commented Apr 9, 2012 at 2:25
  • $\begingroup$ @Argon: you'll want to read A Radical Approach to Real Analysis by Bressoud; that book has a delightful discussion of the vagaries of manipulating series like Grandi's willy-nilly. $\endgroup$
    – J. M. ain't a mathematician
    Commented Apr 10, 2012 at 8:19
  • $\begingroup$ Lovely answer, thanks $\endgroup$
    – Mark K Cowan
    Commented Feb 28, 2015 at 22:59

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