Here is my explanation. It is certainly more mathematical, but there aren't any complicated ideas and hopefully everyone should be able to follow.
We're looking at arbitrary series, something like 1 - 1 + 2 + 3 + 0 + 0 + ... or 1 + 1/2 + 1/4 + 1/8 + 1/16 + ... or 1 + 1/2 + 1/3 + 1/4 + ..., or 1 + 2 + 3 + 4 + 5 + .... If we look at the collection of all of these series, some of them will sum to a finite value in the traditional limit sense (like the first two examples), and some will not (like the last two examples). So we split the series into two groups; those that converge in the traditional sense and those that do not. The ultimate goal is to extend the standard notion of summation to some of those divergent series. Note that those sums still diverge, they do not converge to any finite number in the traditional sense - what we are saying is that even though they do not sum to a value in the standard sense, we can still assign them a number that makes sense in important ways.
So we want to define a function S that assigns a value to series. S should satisfy two important criteria:
- S should agree with the standard limit for series that converge.
- We should be able to multiply a series by a constant, or add two series (component-wise, with no rearrangement), and S should give the 'right' answer. Essentially, S should be a linear function.
Why we care about the first condition should be obvious - there's no point in extending our notion of summation if we lose all the results we had for 'nice' series. The second condition is less clear, but basically, those are things we can do for all convergent series. (The nixing of rearrangement is actually important even when a sequence converges conditionally, so we don't lose much there).
Thinking about these conditions, mathematicians have come up with multiple approaches for assigning values to divergent series. The most famous are Cesaro summation, Abel summation, and Ramanujan summation. These summation techniques have different strengths, so whether or not we can assign a value to a series depends on how strongly divergent that series is. Cesaro summation is the weakest and the simplest, and Ramanujan is stronger, meaning it assigns values to more series, but also more complicated. One cool result is that (under minor technical conditions) the values assigned by these summations agree, so we can really say that there is a unique 'correct' number to assign to these sums.
Cesaro summation looks at the average of the partial sums. If that average converges, then we assign that limit to the original series. So, for example 1 - 1 + 1 - 1 + ... has partial sums 1, 0, 1, 0, 1, ... The average of those numbers approaches 1/2 in the limit, and so we say 1 - 1 + 1 - 1 + ... = 1/2 (not really equal, but it is the correct assignment). Unfortunately, Cesaro summation is not powerful enough to determine the appropriate value for 1 - 2 + 3 - 4 + ... because the partial sums are 1, -1, 2, -2, 3, -3, ... and their average does not converge.
Abel summation notes that we can add a dummy variable to 1 - 2 + 3 - 4 + 5 - ... to make it into a power series 1 - 2x + 3x
2 - 4x
3 + ... This power series has a finite limit for x = 0, and we can increase x to approach 1 to find the appropriate value. So 1 - 1 + 1 - 1 + ... becomes 1 - x + x
2 - x
3+ ... This is a geometric series and sums to 1/(1 + x) for |x| < 1. We then take the limit as x goes to 1, and get 1/2 again. Similarly, 1 - 2x + 3x
2 - 4x
3 + ... = 1/(1 + x)
2, so as x goes to 1 we get 1/4. However, Abel summation is
still not powerful enough to handle the (very divergent) 1 + 2 + 3 + 4 + ...
Ramanujan summation is like Abel summation, but with a more complicated form. We can get a formula for the sum from calculus including derivatives of the function and the Bernoulli numbers. Using that idea, we finally get 1 + 2 + 3 + 4 + ... = -1/12. As mentioned above, another way to approach this is via the analytic continuation of the Riemann zeta function, and the amazing thing is less the value they give (although -1/12 is pretty crazy) but really that they give the same result at all.
