Sounds strange, doesn't it? I'll help prove it.
Let's declare a few series so that we can derive where this comes from. I won't bother to include all the reasons why this works (talking about convergence and whatnot), just so you get the gist. I'll do this first using summation notation, and then I'll show in plaintext what each of them means.
S1(z) = 1 + n=1∑∞{(−1)^n} [(n+1)^(−z) * (−n)^(−z)]
S2(z) = n=1∑∞{(−1)^(n+1) * n^(−z)}
S(z) = n=1∑∞{n^(−z)}
The simply declare some infinite series. I'll write them out below.
S1 = 1 - 1 + 1 - 1 + ...
S2 = 1 - 2 + 3 - 4 + ...
S = 1 + 2 + 3 + 4 + ...
Now that we have these, we can use them to help prove that S(z) is effectively equal to -1/12. The first thing I'm going to do is add the series S2(z) to itself (2S2(z)) so that I can equate it to one of the other series I've previously defined.
2S2 = (1 - 2 + 3 - 4 + ...) + (1 - 2 + 3 - 4 + ...) <-- this is shifted along slightly, i'll make that clear here:
2S2 = (1 - 2 + 3 - 4 + ...)
____+ (0 + 1 - 2 + 3 - 4 + ...)
2S2 = 1 - 1 + 1 - 1 + ...
2S2 = S1
Now we know that 2S2(z) = S1(z). So, let's solve for S1(z).
S1 = 1 - 1 + 1 - 1 + ...
1 - S1 = 1 - (1 - 1 + 1 - 1 + ...)
1 - S1 = 1 - 1 + 1 - 1 + ...
1 - S1 = S1
1 = 2S1
1/2 = S1
Now we know that S1(z) = 1/2, which means that S2(z) = 1/4. Here's where we get to see how 1 + 2 + 3 + 4 + ... infinitely must equal -1/12.
S = 1 + 2 + 3 + 4 + ...
S - S2 = (1 + 2 + 3 + 4 + ...)
____- (1 - 2 + 3 - 4 + ...)
S - S2 = 4 + 8 + 12 + 16 + ...
S - S2 = 4(1 + 2 + 3 + 4 + 5)
S - 1/4 = 4S
-1/4 = 3S
-(1/4)/3 = S
S = -1/12
Pretty cool, eh?
Keep in mind I obviously didn't come up with this myself, this is from a paper by Dr. Tony Padilla. Link here.
