Does this prove that the sum of the positive integers is -1/12?

Right away, S₁ does not have a value as you claim. Your very first "proof" is wrong:

You assume S₁ = x, you then proceed to show that, assuming this:

x = 1 - 1 + 1 - 1 + ...

x - 1 = -1 + 1 - 1 + 1 + ...

Next, you erroneously assume that x + (x - 1) = 0. No, let's look at what it actually equals:

There are two possible cases for the sum:

Case 1:

x = 1 - 1 + 1 - 1 + ... + 1 - 1

x - 1 = 0 - 1 + 1 - 1 + ... 1 - 1

--> summing these two is clearly NOT zero. First, the original sum is clearly x = 0, subtracting 1 obviously gives x - 1 = -1

And you can see this. If you shift the second series down by one, you get:

x - 1 = -1 + 1 - 1 + ... -1 <-- there is one less term at the end

So everything cancels EXCEPT the last -1 in the original (because the last two terms were 1 - 1 <-- the 1 cancels with that last -1 in the second series and leaves -1. Therefore you have:

x + x - 1 = -1 --> 2x = 0 --> x = 0, which is correct (for this case)

Case 2: you have an odd number of terms

x = 1 - 1 + 1 - 1 + ... + 1 - 1 + 1 --> clearly x = 1

x - 1 = -1 + 1 - 1 + ... - 1 + 1

--> similar reasoning to above, everything cancels except the last term in the original series, giving

x + x - 1 = 1 --> 2x = 2 --> x = 1, which is correct for this case

Edit:

I'm not really sure what significance this value has that they claim in the video (where they claim that the summation must simply be half of the two possible values). I don't see what relevance it could have to anything meaningful because mathematically it's not correct. It is incorrect to state that the sum of a divergent series has a value. If you want to define some other number, which is say, the average of all of the partial sums, then fine, we can analyze this value.

This is the number they are getting when they say:

{ Σ(-1)^n, from n = 1 to ∞ } "equals" 1/2 <-- I'm using quotes, because IT DOES NOT EQUAL THAT VALUE!!!

Now, we can define the average of the partial sums. Pretty simple:

{ Σ(-1)^n, from n = 1 to n = N } = ((-1)^N - 1) / 2 <-- so that when N = 1, we get -1 and when N is even, we get (1 - 1)/2 = 0

So now we simply do this new from n = 1 to n = N:

{ Σ(((-1)^n - 1) / 2), from n = 1, to n = N)

well, this is extremely easy to do, because we already have know the sum of (-1)^n, which appears here, it's either 0 or -1. Now when we divide by N (to take the average of the first N partial sums), we will get:

{0, -1}/(2N) - N/(2N) <--the N is from summing -1/2 N times.

This is just {0, -1} / (2N) - 1/2 <-- if we take the limit as N tend towards infinity, then the first part becomes 0 (regardless of whether it's 0 or -1) and thus the final AVERAGE of the partial sums is indeed -1/2.

Which is kind of funny to me because you get the "correct" result (because you start at n = 0 and thus 1 - 1) whereas in the video they start at n = 1 and thus get the incorrect result of +1/2 when it should have been -1/2.

And using this definition, you still get divergence even when asking what the average of the partial sums are of the integers: http://latex.codecogs.com/gif.latex?%5Csum_1%5EN%2...

sum of squares: http://en.wikipedia.org/wiki/Square_pyramidal_numb...

When dealing with infinite sums, expect surprises.