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I was watching this video I noticed that the teacher said that 1-1+1-1+1... equals 1/2. How can we know that? The proof he uses doesn't make sense to me. We go from 1 to 0 to 1 and back again, etc. If it goes on like so forever, where does a fraction come into play?

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    It doesn't make sense. This is one of the better known fake proofs. The sum simply does not exist in the normal sense of sums of infinitely many real numbers, i.e. a series. Or, perhaps more accurately, it cannot be assigned any real number as a value, and the series diverges. – Jyrki Lahtonen Aug 11 '16 at 18:48
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    The series does not converge, because as you observe, the partial sums $1,0,1,0,\ldots$ oscillate and do not approach a single limit. There are various ways to assign values to some divergent series, such as Cesaro summation, regularization, etc. See here for some information specific to this series. –  Aug 11 '16 at 18:48
  • It depends in the definition. It is not conventionally convergent you are right. But notice 1-1/2=1/2 and 1-S=S. Thus 1/2 is the only possible sum. – Jacob Wakem Aug 11 '16 at 18:50
  • Using the traditional methods of summation, you can't. However, the Cesaro sum is the average of the partial sums, and the Cesaro sum exists and equals $\frac 12.$ But the Cesaro sum, isn't the sum. – Doug M Aug 11 '16 at 18:51
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    Duplicate maybe: http://math.stackexchange.com/q/635324/321264, http://math.stackexchange.com/q/648665/321264 – StubbornAtom Aug 11 '16 at 18:52
  • http://karagila.org/2016/syntactic-t-rex-irregularized/ – Asaf Karagila Aug 11 '16 at 18:57

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The series does not converge

But if you treated it as a formal geometric series $1+r+r^2+\ldots= \dfrac{1}{1-r}$ and then let $r=-1$ you would get $\dfrac{1}{2}$

Similarly if $S=1-1+1-1+\cdots$ then you might set up 

 S = 1 - 1 + 1 - 1 + ...  
 S =     1 - 1 + 1 - ...

and adding the two vertically gives

2S = 1
Henry
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It's a non-convergent series (Grandi's Series), so all we can say is that it diverges. Otherwise you get weird results.

For example:

$$S = 1 - 1 + 1 - 1 + 1 - 1 + ... = 1 - (1 -1) - (1 - 1) - (1-1) - ... = 1$$

or

$$S = 1 - 1 + 1 - 1 + 1 - 1 + ... = (1-1) + (1-1) + (1-1) + ... = 0$$

or

$$1 - S = 1 - (1 - 1 + 1 - 1 + 1 - 1 + ...) = S \implies S = 1/2$$

user51819
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That series does not converge, because the limit of the partial sums ($0, 1, 0, 1, ...$) does not exist.

FalafelPita
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There are multiple measures of what that comes to but intuitively you might think that the value alternates between 1 and 0, so you could call it a half.

In truth this series never converges on any given number. Depending on how you define addition, the sum to infinity is not properly defined. You might argue that the sum of those numbers is 1 if $\infty$ is an odd number and $0$ if $\infty$ is an even number.But $\infty$ is neither an even nor an odd number.

Summing to infinity cannot truly be evaluated as a sum as it is a process of addition that never ends, rather than a final result. What we can say fairly categorically is that the only acceptable numerical answers are $1, 0,$ and $\frac{1}{2}$.

it's a hire car baby
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For divergent series, you can use Cesàro summation to find your desired limit.

The idea is to take the mean of your partial sums, and we hope it converges:

1 1 1/1
2 0 1/2
3 1 2/3
4 0 2/4
5 1 3/5

etc. you can see the mean goes to $1/2$.

Kaa1el
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