For n a non-negative integer, $ζ(-n)$ can be interpreted as assigning a value to the (divergent) series $1^n+2^n+3^n+4^n+\cdots$


A value can also be assigned to the related series ${n+0 \choose n}+{n+1 \choose n}+{n+2 \choose n}+{n+3 \choose n}+\cdots$ by comparing its value with that of "powers" of Grandi's series $1-1+1-1+\cdots$


Concretely (but not rigorously):


Considering $1+x+x^2+x^3+x^4+\cdots$ as a formal power series or by analytic continuation, it can be identified with $\frac{1}{1-x}$. This gives a well-defined value except at $x=1$, but then we can use the fact that, where the terms are defined,


$\frac{1}{1-x} - \frac{2x}{1-x^2} = \frac{1}{1+x}$


or, in power series,


$(1+x+x^2+x^3+\cdots)-2x(1+x^2+x^4+x^6+\cdots)=1-x+x^2-x^3+\cdots$


and setting x = 1,


$(1+1+1+1+\cdots) - 2(1+1+1+1+\cdots) = 1-1+1-1+\cdots$


where the right-hand side is already assigned the value of $1/2$, making $1+1+1+1+\cdots = -1/2$. This agrees with the value given by $ζ(0)$.


Similarly,


$\frac{1}{(1-x)^2} - \frac{4x}{(1-x^2)^2} = \frac{1}{(1+x)^2}$


or, in power series,


$(1+2x+3x^2+4x^3+\cdots)-4x(1+2x^2+3x^4+4x^6+\cdots)=1-2x+3x^2-4x^3+\cdots$


and setting x = 1,


$(1+2+3+4+\cdots) - 4(1+2+3+4+\cdots) = 1-2+3-4+\cdots$


where the right-hand side is already assigned the value of $(1/2)^2 = 1/4$, making $1+2+3+4+\cdots = -1/12$. Again, this agrees with the value given by $ζ(-1)$.


This is all extremely hand-wavy. In particular, the argument above only works because we chose to look at ("powers" of) Grandi's series $1-1+1-1+\cdots$ and not some other geometric series $1+y+y^2+y^3+\cdots$ for $y≠-1$.


My question then is: *why* does this work for Grandi's series? Surely it isn't all just a coincidence?



ζ(-n) and "powers" of Grandi's series
[3/30 -- Not sure anyone is paying attention to this any more, but I attempted to fill in the gaps mentioned in the comments]


I was sort of of hoping an expert would come by, but I'll give it a shot.


The short answer is that Grandi's series and its "powers" aren't very divergent. In particular, Tao's smoothed asymptotics have no terms with a positive power of $N$ and give the same answer as evaluating the power series at 1.
$$
\sum (-1)^n \sim \frac{1}{2} + O\left(\frac{1}{N}\right)\ .
$$
(I'm using the sim symbol to indicate the smoothed asymptotics.) Similarly, we have 
$$
\sum n (-1)^n \sim \frac{1}{4} + O\left(\frac{1}{N}\right)\ .
$$
More generally, we have
$$
\sum n^m (-1)^n \sim -\frac{2^{m+1}-1}{m+1}B\_{m+1} + O\left(\frac{1}{N}\right)\ .
$$
These sums are Cesaro summable $(C,k)$ for high enough $k$ and the sum is the same as the Abel sum (i.e., evaluating the power series at 1 -- the proof is in Hardy's book, for example) which can be evaluated in terms of Eulerian numbers. $(C,k)$ summability is also equivalent to Riesz summability (also in Hardy's book) which means that there are no divergent terms with the smoothing $\eta(x)=(1-x)\_{+}^k$. However, I don't think this is sufficient, so a proof of this formula for sufficiently nice $\eta(x)$ is given below.


Moving to the the sums of integer powers, with the smoothed asymptotics, they have the form
$$
C + aN^s + O\left(\frac{1}{N}\right)\ .
$$
Then, as shown in the blog post, C is the value you get via zeta function regularization. What you're doing in your question is manipulating the sums to give something where the divergent terms cancel as in the two "Grandi-like" sums above. Therefore, what remains gives an identity you can solve for C, i.e., the value of the zeta function.


Using the asymptotics in the linked blog above, it's easy to see this happen explicitly:
$$
\sum 1 \sim -\frac{1}{2} + aN + O\left(\frac{1}{N}\right)\ ,
$$
$$
\sum n \sim -\frac{1}{12} + bN^2 + O\left(\frac{1}{N}\right)\ ,
$$
$$
\sum n^2 \sim cN^3 + O\left(\frac{1}{N}\right)\ .
$$
So, starting with
$$
(1 + \dots) - 2(1 + \dots) = 1 - 1 + \dots\ ,
$$
you get
$$
C + aN - 2\left(C + a\frac{N}{2}\right) + O\left(\frac{1}{N}\right) = \frac{1}{2} + O\left(\frac{1}{N}\right)
$$
or
$$
-C = 1/2
$$
confirming that $\zeta(0)=-1/2$. The sum for $\zeta(-1)$ works the same where the $4(N/2)^2$ divergence in the second term cancels the $N^2$ divergence in the first. And you can get $\zeta(-2)=0$ using the identity:
$$
\frac{1+x}{(1-x)^3} - \frac{8x(1+x^2)}{(1-x^2)^3} = \frac{1-x}{(1+x)^3}\ .
$$
As above, to do the general case, you can use the Abel summation:
$$
\sum\_n (-1)^n n^m \sim -\frac{1}{2^{m+1}} \sum (-1)^i \left<{m \atop i}\right> = -\frac{2^{m+1}-1}{m+1}B\_{m+1}\ .
$$
and divide to get the usual answer for the zeta function evaluated at negative integers.


Finally, the promised proof that the smoothed alternating sums of powers give the same answer as Abel summation up to terms that vanish as $N\to\infty$.


First, generalizing a statement in Tao's blog post, it's easy to show that:
$$
\sum (-1)^n f(n) = \frac{1}{2^{k-1}} \left((-1)^k \sum (\Delta^k f)(2n) + \sum\_{l=0}^{k-1} (-1)^l f(l) \sum\_{i=1}^{\infty} \left({k \atop l + 2i}\right)\right)\ .
$$
For $k$ large enough and with $\eta(x)$ highly differentiable, supported on $[0,1]$ and $\eta(0)=1$, we can apply this to $f(n) = \eta(n/N)(-1)^n n^m$. Given the assumptions on $k$ and $\eta(x)$, the first term will always go as some negative power of $N$, so all that remains to be shown is that the second sum gives the same answer as the Abelian sum up to terms of order $1/N$. This can be done through a brute force calculation. Here's a quick sketch (assuming I didn't miss anything):


(1) Since $\eta(0)=1$, $\eta(l/N) = 1 + O(1/N)$, so it can be set to 1.  

(2) Let the limits of $l$ go up to $k$ and $i$ down to zero.  

(3) Replace $2i$ by $i$ by multiplying the sum by $(-1)^i + 1$.  

(4) Change variables so $l+i$ is one variable and $l$ is the other.  

(5) One of the sums will go to zero for large enough $k$ because it is a high order finite difference of a lower degree polynomial.  

(6) For the other sum, use the explicit formula for the sum
$$
\sum\_{n=0}^k (-1)^n n^m
$$
in terms of Eulerian numbers.  

(7) More things vanish because they end up being high order finite differences.  

(8) End up with the same alternating sum of Eulerian numbers as in the Abel summation above. 


