# Why 1-1+1-1+1-… could equal to 1/2!

Hi,
this post is made according to Numberphile video which is pretty cool and you can check it out here.

Lets jump right to the problem of sum: 1-1+1-1+1-1+…

The problem with this sum is that it does not converge into some value like this sum:

1+1/2+1/4+1/8+1/16+…=2

We know this because for infinite amount of adding we get 2. The problem appears when you have sum like the one above which does not converge at all but rather oscilates between 0,1,0,1,0,1,0,1…

To get the solution we can do few things, we can give brackets there:

(1-1)+(1-1)+(1-1)+… this gives us 0 because all the brackets equal to 0.

1+(-1+1)+(-1+1)+(-1+1)+… this equals to 1 because all the brackets are 0 but there is this 1 on the start.

There is third solution:

S=1-1+1-1+1-1+1-…

1-S=1-(1-1+1-1+1-1+1-…)=

=1-S=1-1+1-1+1-…

=1-S=S

=2S=1

=S=1/2

In the video they do not say much more about it but I found that this is a method called Cesàro summation. This is just a method so yes you can say that this sum equals to 1/2 according to the method.

In math stack exchange they say this. And on wikipedia they say that Grandi’s series (1-1+1-1+1…) can not be solved like this:

“It can be shown that it is not valid to perform many seemingly innocuous operations on a series, such as reordering individual terms, unless the series is absolutely convergent.”

Though this is probably that much I can say for this sum since I do not yet understand the higher mathematics involved in all those proofs.

Dragallur