4
$\begingroup$

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.

  • 0
    It diverges but can be summed up by various methods. The sum is 1/2.2017-07-05

1 Answers 1

11

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.

  • 0
    Lovely answer, thanks2015-02-28