Whoops, I suppose there are actually existing systematic techniques called “zeta function regularization” in physics; I was just taking the term to mean a sort of loosey-goosey way of saying “I’ll sum this series by pretending it came from the Riemann zeta function in some ad hoc way”, and pointing out one way of making such things rigorous.
The reason I don’t like interleaving with zeros and claiming it doesn’t change the answer is that you immediately get different answers for arguably the simplest case:
X = 1 - 1 + 1 - 1 + ...
X = 1 - 1 + 1 - 1 + ...
2X = 1 + 0 + 0 + 0 + ...
2X = 1
But
X = 1 + 0 - 1 + 1 + 0 - 1 + ...
X = 1 + 0 - 1 + 1 + 0 - 1 + ...
X = 1 + 0 - 1 + 1 + 0 - 1 + ...
3X = 1 + 1 + 0 + 0 + 0 + ...
3X = 2
I’m sorry; I miscalculated. The bolded value here should be 5/12.
Some convenient properties for series manipulation in terms of the associated function as defined above: If the original series has associated function f(h), then:
The series shifted “down” [so the term which was at index i goes to index i + 1] has associated function f(h) e[sup]h[/sup]. [This can change the sum, but won’t if the sum already exists at degree 0]
Moving the original series to the indices which are multiples of k while interleaving zeros elsewhere has function f(kh) [which will not change the sum at any degree, though other kinds of interleaving amount to this mixed with potentially sum-changing shifts].
Multiplying each term by its index corresponds to differentiating the associated function; if the old series was summable at degree d, the new series will be summable at degree d + 1.
Multiplying each term by the pth power of its index has the associated function f(h - log(p)).
Also, perhaps too trivially to mention, a linear combination of series induces the same linear combination of the corresponding functions, and a finite series [i.e., one which is 0 from some point on] is degree summable to the usual value.
Thus, revisiting the naive reasoning paying attention to what this particular summation method does and does not justify:
Note that all series in this post are being taken as starting at index 1 unless otherwise indicated.
It’s not actually immediate that X and X shifted over by one position have the same sum; however, if we knew that X was degree 0 summable, this would be legitimate.
Being careful about shifts, all this reasoning actually establishes is that the function associated to X satisfies f[sub]X/sub + f[sub]X/sube[sub]h[/sub] = the function associated to 1 + 0 + 0 + 0 + … . That is, we have that f[sub]X/sub = the function associated to 1[sub]1[/sub]/(the function associated to 1[sub]0[/sub] + 1[sub]1[/sub]), where the subscripts explicitly indicate corresponding indices.
But degree 0 summability being merely a limit, and limits respecting division, all that it takes to justify this derivation of X’s value, on this summation method, is the observation that finite sums are degree 0 summable. [Explicitly, we have that f[sub]X[/sub] = e[sup]h[/sup]/(1 + e[sup]h[/sup]), where the numerator and denominator both have limits at h = 0, and thus so does f[sub]X[/sub]]
This is justified in exactly the same way as before; although shifts needn’t generally match, we have that f[sub]Y[/sub] = f[sub]X[/sub]/(1 + e[sup]h[/sup]) and thus, as X is degree 0 summable, so is Y, to X/2.
Again, the only concern is that Y and Y shifted over by one position needn’t match up unless Y is degree 0 summable. But Y is in fact degree 0 summable (this reasoning shows that Y’s associated function is X’s associated function/(1 + e[sup]h[/sup]) = e[sup]h[/sup]/(1 + e[sup]h[/sup])[sup]2[/sup]).
Note that the interleaving here is of the kind which places the original series at the even indices; thus, the interleaving does not change the sum, and this reasoning is justified on our summation method as long as we know that Z is summable at some degree. Which it is, although this argument has not quite established that lemma.
[As f[sub]Z[/sub] is the derivative of the series associated to 1 + 1 + 1 + 1 + …, we will know that Z is degree 2 summable once we know that 1 + 1 + 1 + 1 + … is degree 1 summable. This argument doesn’t establish the latter, but, for what it’s worth, by the obvious shifting recurrence, we see that 1 + 1 + 1 + 1 + … has associated series satisfying f(h) = e[sup]h[/sup] + e[sup]h[/sup]f(h); solving for f(h), we get e[sup]h[/sup]/(1 - e[sup]h[/sup]). To establish degree 1 summability, we multiply by h, differentiate, then take the limit at h = 0; this is tedious, but will yield -1/2, letting the above argument then do the work of showing Z = -1/12.]
And, as we’ve seen, this particular summation method is one which does justify this reasoning, at least once we know that 1 + 2 + 3 + 4 + … is summable at all.
We can also now explain
Recall that Z shifted over needn’t have the same value as Z, as Z is not degree 0 summable. And if we take all these series to be starting at index 1, the interleaving here puts the zeros at the even indices rather than the odd indices; thus, it needn’t preserve value either. [We can calculate this interleaving’s effect on associated functions by the fact that it amounts to shifting down, interleaving zeros into the odd indices, and then shifting back up, but those shifts can cause problems for this reasoning].
What we have is that:
Z shifted over is degree 1 summable to 5/12
Z + Z shifted over is degree 1 summable to 1/3 (which = -1/12 + 5/12, as it must be)
Z + Z shifted over interspersed with zeros at even indices is degree 2 summable to 1/12.
If instead we take everything to be starting at index 0, at least the interspersing would work nicely in your argument.
Thus, your argument also shows that:
Z shifted back + Y shifted back = 2(Z shifted back + Z), which would work out as:
Y shifted back = Y = 1/4 [as Y is degree 0 summable, and thus shift invariant]
Z = -1/12 [as shown above]
Z shifted back = 5/12 [this happens to equal Z shifted forward, as Z shifted back - Z shifted forward = 1 + 2 + 2 + 2 + 2 + … starting from index 0 = 1 + 2(-1/2) = 0; the analogue of this reasoning shows more generally that Z shifted forward by n indices or backwards by n indices produces the same result]
And 5/12 + 1/4 does indeed equal 2(5/12 + -1/12).
So it’s all consistent, within its own rules. It’s unfortunate that shifting is not value-preserving for series which are only summable at positive degree. But that’s how it goes…
Tackling this as well using this particular summability method:
Let’s call the top argument here a description of X and the bottom argument a description of X’.
The top argument here does show that X = 1/2, the shift causing no problems in this case as explained before, and the bottom argument in an analogous way showing that X’ = 2/3, both of these sums being at degree 0.
Now, X does not equal X’, but that’s because the interspersing here is of a tricky sort (remember, the only interspersing that automatically works nicely on this summation method is when we multiply all the indices of the original series through by the same value).
For convenience, I’ll take everything here to start from index 0. Let U = 1 + 1 + 1 + 1 + … We have that:
The interspersing relationship between X and X’ is described by how X and X’ separately relate to U.
Breaking X into its 1 and -1 components gives its relationship to U:
We have that f[sub]X/sub = (1 - e[sup]h[/sup])f[sub]U/sub.
Breaking X’ into its 1, 0, and -1 components gives its relationship to U:
f[sub]X’/sub = (1 - e[sup]2h[/sup])f[sub]U/sub.
And the arguments above demonstrate that f[sub]X[/sub] = 1/(1 + e[sup]h[/sup]), while f[sub]X’[/sub] = (1 + e[sup]h[/sup])/(1 + e[sup]h[/sup] + e[sup]2h[/sup]).
Which is consistent with the above, given that f[sub]U[/sub] = 1/(1 - e[sup]h[/sup]). It fits exactly as the algebra demands.
[Note: You might think, because X and X’ are degree 0 summable, that we should be able to plug in h = 0 and the value of U = 1/2 to the above expressions to evaluate both X and X’ (getting zero for both). However, this will not work as U is not degree 0 summable, and multiplication of associated functions does not play as nicely with positive degree summation as you might expect]
Ha! You made a mistake. Now we can ignore everything you say!
Wait. My mistake. I thought this was a political thread. 
There’s actually a much nicer way to see that this should sum to -1/2, but first, we’ll need to extend our summation method so it handles not only series which extend infinitely to the right, but also those which extend infinitely to the left (i.e., into negative indices).
This is easy enough: recall, that given a series {a[sub]i[/sub]} to sum, we construct an associated function f(h) = sum {a[sub]i[/sub] e[sup]-ih[/sup]}, and then proceeded to extrapolate a value for f at h = 0. If our series extended infinitely only to the right, we took a limit as h approached 0 from the right (as such a series is more likely to converge for more positive values of h); analogously, for a series which extends infinitely to the left, we shall take the same limit as h approaches 0 from the left (as such a series is more likely to converge for more negative values of h). For a general series which may extend infinitely in both directions, we may simply split it into two pieces, evaluate them separately, and add the results together [it’s easy enough to see that the choice of where to split makes no difference].
This generalization preserves essentially all the properties of associated function manipulation we’ve been using so far, and also has the natural property that flipping a series [negating all the indices] doesn’t change its sum.
With this generalization in mind, consider the series …1 + 1 + 1 + 1 + … = (the series of 1s at every integer index). On the one hand, this is unchanged by a shift forward, so we have that its associated function satisfies f(h) = f(h)e[sup]h[/sup], which is to say, f(h) = 0/(1 - e[sup]h[/sup]) = 0, and thus this series sums to 0.
On the other hand, we also have that (the series of 1s at every integer index) = 1 + (the series of 1s at positive indices) + (the series of 1s at negative indices) = 1 + 2(the series of 1s at positive indices). Thus, this must also equal 0, and we readily see that the series of 1s at positive indices sums to -1/2.
Finally (I think), there’s one other improvement we can make to our summation method: currently, this summation method can only handle series {a[sub]i[/sub]} for which the associated power series {a[sub]i[/sub]X[sup]i[/sup]} has a radius of convergence of at least 1 [as otherwise, the associated function f(h) will not be defined for h near 0]. This prevents us from summing 1 + 2 + 4 + 8 + …, 1 + 1 + 2 + 3 + 5 + …, and the like.
But it is easily remedied by a further generalization: For any series, its associated function either converges nowhere [in which case, we won’t attempt to assign a sum], or converges for sufficiently large h, on which it defines an analytic function (essentially, a “nice” or “smooth” function). It may be that this function can be extended to a ratio of analytic functions on all positive h, and if so, this extension is unique. We can then freely identify the associated function with its extension, and set up the limits defining summation under our method as before.
Thus, even though the associated function for 1 + 2 + 4 + 8 + … (i.e., 1e[sup]-1h[/sup] + 2e[sup]-2h[/sup] + 4e[sup]-4h[/sup] + 8e[sup]-8h[/sup] + …) converges to 1/(1 - 2e[sup]-h[/sup]) only for h > ln(2) and fails to converge elsewhere, we can happily reason as though it were defined by this expression on all positive h, and find that 1 + 2 + 4 + 8 + … is degree 0 summable to -1.
[Yes, I’ve had to invoke analytic continuation here (sigh), but still in a systematic manner, rather than an ad hoc one; this summation method remains a deterministic process of assigning a value to a given series, not depending on where the series “came from” and only depending on what terms are at which indices…]