We start from the simpler question of what is 1 - 1 + 1 - 1 + 1 - 1 + ... for an infinite number of terms. If you stop the series at a 1 then the answer is 1 but if you stop at a -1 then the answer is 0. Since the series never stops then we can make the assumption that the answer is 0.5, which is the average. This is like if you had a light switch that you were constantly switching on and off. If done at a fast enough frequency, the light will seem to be halfway light and dark, which is 0.5.

$$\sum_i{(-1)^i} = \frac{1}{2}$$

If you accept the above assumption then the rest is rigorous.

We next need to find what 1 - 2 + 3 - 4 + 5 - ... is. This can be reduced to the above identity by applying the following trick:

S = 1 - 2 + 3 - 4 + 5 - ... S+S = 1 - 2 + 3 - 4 + 5 - ... + 1 - 2 + 3 - 4 + ... = 1 - 1 + 1 - 1 + 1 - ...

Therefore $2S = \frac{1}{2}$ which means that $S = \frac{1}{4}$.

$$\sum_i{(-1)^i i} = \frac{1}{4}$$

Finally we get to our original question:

S' = 1 + 2 + 3 + 4 + 5 + ... S'-S = 1 + 2 + 3 + 4 + 5 + 6 + ... - 1 + 2 - 3 + 4 - 5 + 6 - ... = 0 + 4 + 0 + 8 + 0 + 12 + ...

Which are the even numbers multiplied by 2, that is, the multiples of 4.

S'-S = 4 + 8 + 12 + ... = 4(1 + 2 + 3 + ...) = 4S'

Now if S' - S = 4S' then

S' - S = 4S' S' - 4S' = S -3S' = S S' = -S/3 = -(1/4)/3 = -1/12

$$\sum_i{i} = -\frac{1}{12}$$